Question 14 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\big(\frac\pi2-\text{x}\big)\sin\text{x}-2\cos\text{x}}{\big(\frac\pi2-\text{x}\big)+\cot\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\big(\frac\pi2-\text{x}\big)\sin\text{x}-2\cos\text{x}}{\big(\frac\pi2-\text{x}\big)+\cot\text{x}}$ If $\text{x}\rightarrow\frac{\pi}{2},$ then $\frac\pi2-\text{x}\rightarrow0$ let $\frac\pi2-\text{x}=\text{y}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\Big(\text{y}\sin\big(\frac{\pi}{2}-\text{y}\big)-2\cos\big(\frac{\pi}{2}-\text{y}\big)\Big)}{\text{y}+\cot\big(\frac\pi2-\text{y}\big)}$ $=\lim\limits_{\text{y}\rightarrow{0}}\Big(\frac{\text{y}\cos\text{y}-2\sin\text{y}}{1+\tan\text{y}}\Big)$ $=\lim\limits_{\text{y}\rightarrow{0}}\Bigg(\frac{\cos\text{y}-2\frac{\sin\text{y}}{\text{y}}}{1+\frac{\tan\text{y}}{\text{y}}}\Bigg)$ $=\frac{\lim\limits_{\text{y}\rightarrow{0}}\cos\text{y}-2\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\text{y}}{\text{y}}}{1+\lim\limits_{\text{y}\rightarrow{0}}\frac{\tan\text{y}}{\text{y}}}$ $\Big[\because\lim\limits_{\text{y}\rightarrow0}\frac{\sin\text{y}}{\text{y}}=1,\lim\limits_{\text{y}\rightarrow0}\frac{\tan\text{y}}{\text{y}}=1\Big]$ $=\frac{1-2}{1+1}=\frac{-1}{2}$ $=-\frac12$
View full question & answer→Question 24 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow5}\frac{\text{x}-5}{\sqrt{6\text{x}-5}-\sqrt{4\text{x}+5}}$
Answer$\lim\limits_{\text{x}\rightarrow5}\frac{\text{x}-5}{\sqrt{6\text{x}-5}-\sqrt{4\text{x}+5}}$ $=\lim\limits_{\text{x}\rightarrow5}\frac{\text{x}-5}{\big(\sqrt{6\text{x}-5}-\sqrt{4\text{x}+5}\big)}\times\frac{\big(\sqrt{6\text{x}-5}+\sqrt{4\text{x}+5}\big)}{\big(\sqrt{6\text{x}-5}+\sqrt{4\text{x}+5}\big)}$ $=\lim\limits_{\text{x}\rightarrow5}\frac{(\text{x}-5)\big(\sqrt{6\text{x}-5}+\sqrt{4\text{x}+5}\big)}{(6\text{x}-5)-(4\text{x}+5)}$ $=\lim\limits_{\text{x}\rightarrow5}\frac{(\text{x}-5)\big(\sqrt{6\text{x}-5}+\sqrt{4\text{x}+5}\big)}{2\text{x}-10}$ $=\lim\limits_{\text{x}\rightarrow5}\frac{(\text{x}-5)\big(\sqrt{6\text{x}-5}+\sqrt{4\text{x}+5}\big)}{2(\text{x}-5)}$ $=\frac{\sqrt{6(5)-5}+\sqrt{4(5)+5}}{2}$ $=\frac{\sqrt{25}+\sqrt{25}}{2}$ $=\frac{5+5}{2}=5$
View full question & answer→Question 34 Marks
Evaluate the following limit: $\lim\limits_{\text{h}\rightarrow0}\frac{(\text{a}+\text{h})^2\sin(\text{a}+\text{h})-\text{a}^2\sin\text{a}}{\text{h}}$
Answer$\lim\limits_{\text{h}\rightarrow0}\frac{(\text{a}+\text{h})^2\sin(\text{a}+\text{h})-\text{a}^2\sin\text{a}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{a}+\text{h})^2(\sin\text{a}\cos\text{h})-\text{a}^2\sin\text{a}+(\text{a}+\text{h})^2\cos\text{a}\sin\text{h}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\big(\text{a}^2+2\text{ah}+\text{h}\big)(\sin\text{a}\cos\text{h})-\text{a}^2\sin\text{a}+(\text{a}+\text{h})^2\cos\text{a}\sin\text{h}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}^2\sin\text{a}(\cos-1)+2\text{ah}\sin\text{a}\cos\text{h}+\text{h}^2\sin\text{a}\cos\text{h}+(\text{a}+\text{h})^2\cos\text{a}\sin\text{h}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}^2\sin\text{a}(\cos-1)}{\text{h}}+\lim\limits_{\text{h}\rightarrow0}\frac{2\text{ah}\sin\text{a}\cos\text{h}}{\text{h}}\\\ +\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}^2\sin\text{a}\cos\text{h}}{\text{h}}+\lim\limits_{\text{h}\rightarrow0}\frac{(\text{a}+\text{h})^2\sin\text{a}\cos\text{h}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{-\text{a}^2\sin\text{a}\sin^2\big(\frac{\text{h}}{2}\big)}{\frac{\text{h}}{2}}+2\text{a}\sin\text{a}+0+\text{a}^2\cos\text{a}$ $=0+2\text{a}\sin\text{a}+\text{a}^2\cos\text{a}$ $=2\text{a}\sin\text{a}+\text{a}^2\cos\text{a}$
View full question & answer→Question 44 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\{\sin(\alpha+\beta)\text{x}+\sin(\alpha-\beta)\text{x}+\sin2\alpha\text{x}\}}{\cos^2\beta\text{x}-\cos^2\alpha\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\{\sin(\alpha+\beta)\text{x}+\sin(\alpha-\beta)\text{x}+\sin2\alpha\text{x}\}}{\cos^2\beta\text{x}-\cos^2\alpha\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\Big\{2\sin\frac{(\alpha+\beta+\alpha-\beta)}{2}\times\cos\frac{(\alpha+\beta-\alpha+\beta)}{2}\times+2\sin\alpha\cos\alpha\text{x}\Big\}}{(\cos\beta\text{x}-\cos\alpha\text{x})(\cos\beta\text{x}+\cos\alpha\text{x})}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\big\{2\sin\alpha\text{x}\cos\beta\text{x}+2\sin\alpha\text{x}\cos\alpha\text{x}\big\}}{(\cos\beta\text{x}-\cos\alpha\text{x})(\cos\beta\text{x}+\cos\alpha\text{x})}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\alpha\text{x}(\cos\beta\text{x}+\cos\alpha\text{x})}{(\cos\beta\text{x}-\cos\alpha\text{x})(\cos\beta\text{x}+\cos\alpha\text{x})}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\alpha\text{x}}{(\cos\beta\text{x}-\cos\alpha\text{x})}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\alpha\text{x}}{\Big(1-2\sin^2\big(\frac{\beta\text{x}}{2}\big)-1+2\sin^2\big(\frac{\alpha\text{x}}{2}\big)\Big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\alpha\text{x}}{2\sin^2\big(\frac{\alpha\text{x}}{2}\big)-2\sin^2\big(\frac{\beta\text{x}}{2}\big)}$ $=\frac{2\alpha}{\alpha^2-\beta^2}$
View full question & answer→Question 54 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{\sqrt{5+2\text{x}}-\big(\sqrt{3}+\sqrt{2}\big)}{\text{x}^2-6}$
Answer$\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{\sqrt{5+2\text{x}}-\big(\sqrt{3}+\sqrt{2}\big)}{\text{x}^2-6}$$=\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{\sqrt{5+2\text{x}}-\Big(\sqrt{\big(\sqrt{3}+\sqrt{2}\big)^2}\Big)}{\text{x}^2-6}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{\sqrt{5+2\text{x}}-\big(\sqrt{5+2\sqrt{6}}\big)}{\text{x}^2-6}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{\sqrt{5+2\text{x}}-\big(\sqrt{5+2\sqrt{6}}\big)}{\text{x}^2-6}\times\frac{\sqrt{5+2\text{x}}+\big(\sqrt{5+2\sqrt{6}}\big)}{\sqrt{5+2\text{x}}+\big(\sqrt{5+2\sqrt{6}}\big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{{7+2\text{x}}-{7-2\sqrt{10}}}{\big(\text{x}^2-10\big)\Big(\sqrt{7+2\text{x}}+\big(\sqrt{7+2\sqrt{10}}\big)\Big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{5+2\text{x}-5-2\sqrt{6}}{\big(\text{x}^2-6\big)\Big(\sqrt{5+2\text{x}}+\big(\sqrt{5+2\sqrt{6}}\big)\Big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{2\big(\text{x}-\sqrt{6}\big)}{\big(\text{x}^2-\sqrt{6}\big)\Big(\sqrt{5+2\text{x}}+\big(\sqrt{5+2\sqrt{10}}\big)\Big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{2}{\big(\text{x}+\sqrt{6}\big)\Big(\sqrt{5+2\text{x}}+\big(\sqrt{5+2\sqrt{6}}\big)\Big)}$
$=\frac{2}{\big(2\sqrt{6}\big)\Big(2\sqrt{5+2\sqrt{6}}\Big)}$
$=\frac{1}{\big(2\sqrt{6}\big)\Big(\sqrt{5+2\sqrt{6}}\Big)}$
$=\frac{1}{\big(2\sqrt{6}\big)\Big(\sqrt{3}+\sqrt{2}\Big)}$
$=\frac{\big(\sqrt{3}-\sqrt{2}\big)}{\big(2\sqrt{6}\big)}$
View full question & answer→Question 64 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{2-\text{cosec}^2\text{x}}{1-\cot\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{2-\text{cosec}^2\text{x}}{1-\cot\text{x}}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{2-\big(1+\cot^2\text{x}\big)}{1-\cot\text{x}}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{2-1-\cot^2\text{x}}{1-\cot\text{x}}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{1-\cot^2\text{x}}{1-\cot\text{x}}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{(1-\cot\text{x})(1+\cot\text{x})}{(1-\cot\text{x})}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}(1+\cot\text{x})$ $=1+\cot\frac{\pi}{4}$ $=1+1$ $=2$
View full question & answer→Question 74 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+3\text{x}}-\sqrt{1-3\text{x}}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+3\text{x}}-\sqrt{1-3\text{x}}}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+3\text{x}}-\sqrt{1-3\text{x}}\big)}{\text{x}}\times\frac{\big(\sqrt{1+3\text{x}}+\sqrt{1-3\text{x}}\big)}{\big(\sqrt{1+3\text{x}}+\sqrt{1-3\text{x}}\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{(1+3\text{x})-(1-3\text{x})}{\big(\sqrt{1+3\text{x}}+\sqrt{1-3\text{x}}\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{6\text{x}}{\big(\sqrt{1+3\text{x}}+\sqrt{1-3\text{x}}\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{6}{\big(\sqrt{1+3\text{x}}+\sqrt{1-3\text{x}}\big)}$ $=\frac{6}{\sqrt{1}+\sqrt{1}}$ $=\frac62$ $=3$
View full question & answer→Question 84 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow\pi}\frac{1+\cos\text{x}}{\tan^2\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow\pi}\frac{1+\cos\text{x}}{\tan^2\text{x}}$ As$\text{ x}\rightarrow\pi,\text{x}-\pi\rightarrow0,$let $\text{ x }-\pi=\text{y}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{1+\cos(\pi+\text{y})}{\tan^2(\pi+\text{y})}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{1-\cos\text{y}}{\tan^2\text{y}}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{2\sin^2\frac{\text{y}}{2}}{\tan^2\text{y}}$ $=\frac{\lim\limits_{\text{y}\rightarrow0}2\sin^2\frac{\text{y}}{2}}{{\lim\limits_{\text{y}\rightarrow0}\tan^2\text{y}}}$ $=\frac{2\Bigg(\lim\limits_{\text{y}\rightarrow0}\frac{\frac{\sin\text{y}}{2}}{\frac{\text{y}}{2}}\Bigg)^2\times\frac{\text{y}^2}{4}}{\bigg(\lim\limits_{\text{y}\rightarrow0}\frac{\tan\text{y}}{\text{y}}\bigg)\times\text{y}^2}$ $=\frac{2\times1\times\frac{\text{y}^2}{4}}{1\times\text{y}^2}$ $\begin{bmatrix}\therefore\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\\\lim\limits_{\theta\rightarrow0}\frac{\tan\theta}{\theta}=1 \end{bmatrix}$ $=2\times1\times\frac{1}{4}$ $=\frac{1}{2}$
View full question & answer→Question 94 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{{\pi}}}\frac{1-\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}2\big(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{4}\big)}$
Answer$\lim\limits_{\text{x}\rightarrow{{\pi}}}\frac{1-\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}2\big(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{4}\big)}=\lim\limits_{\text{h}\rightarrow{0}}\frac{1\sin\big(\frac{\pi+\text{h}}{2}\big)}{\cos\big(\frac{\pi+\text{h}}{2}\big)\big(\cos\big(\frac{\pi+\text{h}}{2}\big)-\sin\big(\frac{\pi+\text{h}}{2}\big)\big)}$ $=\lim\limits_{\text{h}\rightarrow{0}}\frac{1-\cos\big(\frac{\text{h}}{2}\big)}{-\sin\big(\frac{\text{h}}{2}\big)\big(\frac{1}{\sqrt{2}}\cos\big(\frac{\text{h}}{4}\big)-\frac{1}{\sqrt{2}}\sin\big(\frac{\text{h}}{4}\big)-\frac{1}{\sqrt{2}}\sin\big(\frac{\text{h}}{4}\big)-\frac{1}{\sqrt{2}}\cos\big(\frac{\text{h}}{4}\big)\big)}$ $=\lim\limits_{\text{h}\rightarrow{0}}\frac{1-\cos\big(\frac{\text{h}}{2}\big)}{\sqrt{2}\sin\big(\frac{\text{h}}{2}\big)\sin\big(\frac{\text{h}}{4}\big)}$ $=\lim\limits_{\text{h}\rightarrow{0}}\frac{2-\sin^2\big(\frac{\text{h}}{4}\big)}{\sqrt{2}\sin\big(\frac{\text{h}}{2}\big)\sin\big(\frac{\text{h}}{4}\big)}$ $=\sqrt{2}\lim\limits_{\text{h}\rightarrow{0}}\frac{\sin\big(\frac{\text{h}}{4}\big)}{\sin\big(\frac{\text{h}}{2}\big)}$ $=\sqrt{2}\lim\limits_{\text{h}\rightarrow{0}}\frac{\frac{\sin\big(\frac{\text{h}}{4}\big)}{\big(\frac{\text{h}}{4}\big)}\times\big(\frac{\text{h}}{4}\big)}{\frac{\sin\big(\frac{\text{h}}{2}\big)}{\big(\frac{\text{h}}{2}\big)}\times\big(\frac{\text{h}}{2}\big)}$ $=\sqrt{2}\times\frac{\frac14}{\frac12}$ $=\frac{1}{\sqrt{2}}$
View full question & answer→Question 104 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{\sqrt{7+2\text{x}}-\big(\sqrt{5}+\sqrt{2}\big)}{\text{x}^2-10}$
Answer$\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{\sqrt{7+2\text{x}}-\big(\sqrt{5}+\sqrt{2}\big)}{\text{x}^2-10}$ $=\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{\sqrt{7+2\text{x}}-\Big(\sqrt{\big(\sqrt{5}+\sqrt{2}\big)^2}\Big)}{\text{x}^2-10}$ $=\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{\sqrt{7+2\text{x}}-\big(\sqrt{7+2\sqrt{10}}\big)}{\text{x}^2-10}$ $=\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{\sqrt{7+2\text{x}}-\big(\sqrt{7+2\sqrt{10}}\big)}{\text{x}^2-10}\times\frac{\sqrt{7+2\text{x}}+\big(\sqrt{7+2\sqrt{10}}\big)}{\sqrt{7+2\text{x}}+\big(\sqrt{7+2\sqrt{10}}\big)}$ $=\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{{7+2\text{x}}-{7-2\sqrt{10}}}{\big(\text{x}^2-10\big)\Big(\sqrt{7+2\text{x}}+\big(\sqrt{7+2\sqrt{10}}\big)\Big)}$ $=\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{2\big(\text{x}-\sqrt{10}\big)}{\big(\text{x}^2-10\big)\Big(\sqrt{7+2\text{x}}+\big(\sqrt{7+2\sqrt{10}}\big)\Big)}$ $=\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{2}{\big(\text{x}-\sqrt{10}\big)\Big(\sqrt{7+2\text{x}}+\big(\sqrt{7+2\sqrt{10}}\big)\Big)}$ $=\frac{2}{\big(\sqrt{10}+\sqrt{10}\big)\Big(\sqrt{7+2\text{x}}+\big(\sqrt{7+2\sqrt{10}}\big)\Big)}$ $=\frac{2}{\big(2\sqrt{10}\big)\Big(2\sqrt{7+2\sqrt{10}}\Big)}$ $=\frac{1}{\big(2\sqrt{10}\big)\Big(\sqrt{7+2\sqrt{10}}\Big)}$ $=\frac{1}{\big(2\sqrt{10}\big)\Big(\sqrt{5}+\sqrt{2}\Big)}$ $=\frac{\big(\sqrt{5}-\sqrt{2}\big)}{\big(6\sqrt{10}\big)}$
View full question & answer→Question 114 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\bigg\{\frac{\text{x}-2}{\text{x}^2-\text{x}}-\frac{1}{\text{x}^2-3\text{x}^2+2\text{x}}\bigg\}$
Answer$\lim\limits_{\text{x}\rightarrow1}\bigg\{\frac{\text{x}-2}{\text{x}^2-\text{x}}-\frac{1}{\text{x}^2-3\text{x}^2+2\text{x}}\bigg\}$$=\lim\limits_{\text{x}\rightarrow1}\Bigg\{\frac{\text{x}-2}{\text{x}(\text{x}-1)}-\frac{1)}{\text{x}\big(\text{x}^2-3\text{x}+2\big)}\Bigg\}$
$= \lim\limits_{\text{x}\rightarrow1}\Bigg\{\frac{\text{x}-2}{\text{x}(\text{x}-1)}-\frac{1}{\text{x}\big(\text{x}^2-3\text{x}+2\big)}\Bigg\}$
$= \lim\limits_{\text{x}\rightarrow1}\Bigg\{\frac{\text{x}-2}{\text{x}(\text{x}-1)}-\frac{1}{\text{x}\big(\text{x}^2-1\text{x}-2\text{x}+2\big)}\Bigg\}$
$= \lim\limits_{\text{x}\rightarrow1}\Bigg\{\frac{\text{x}-2}{\text{x}(\text{x}-1)}-\frac{1}{\text{x}(\text{x}-1)(\text{x}-2)}\Bigg\}$
$=\lim\limits_{\text{x}\rightarrow1}\bigg\{\frac{(\text{x}-2)^2-1}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow1}\bigg\{\frac{\text{x}^2+4-4\text{x}-1}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow1}\bigg\{\frac{\text{x}^2-4\text{x}+3}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow1}\bigg\{\frac{\text{x}^2-4\text{x}+3}{\text{x}(\text{x}-2)(\text{x}-1)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow1}\Big[\frac{\text{x}^2-\text{x}-3\text{x}+3}{\text{x}(\text{x}-1)(\text{x}-2)}\Big]$
$=\lim\limits_{\text{x}\rightarrow1}\Big[\frac{\text{x}(\text{x}-1)-3(\text{x}-1)}{\text{x}(\text{x}-1)(\text{x}-2)}\Big]$
$=\lim\limits_{\text{x}\rightarrow1}\Big[\frac{(\text{x}-3)(\text{x}-1)}{\text{x}(\text{x}-1)(\text{x}-2)}\Big]$
$=\frac{(1-3)}{1(1-2)}$
$=\frac{-2}{-1}$
$=2$
View full question & answer→Question 124 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^2-\text{x}-6}{\text{x}^3+3\text{x}^2+\text{x}-3}$
Answer$\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^2-\text{x}-6}{\text{x}^3+3\text{x}^2+\text{x}-3}$Now $\text{x}^2-\text{x}-6$
$=\text{x}^2-3\text{x}+2\text{x}-6$
$=\text{x}(\text{x}-3)+2(\text{x}-3)$
$=(\text{x}+2)(\text{x}-3)\ \cdots(\text{i})$
Dividing $\text{x}^3-3\text{x}^2+\text{x}-3\text{ by }(\text{x}-3), \text{ we get}$
Thus (x - 3) is a factor of $\text{x}^3-3\text{x}^2+\text{x}-3\ \cdots(\text{ii})$
Substituting (i) and (ii) in the given expression
$=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}+2)(\text{x}-3)}{\big(\text{x}^2+1\big)(\text{x}-3)}$
$=\frac{\text{x}+2}{\text{x}^2+1}=\frac{3+2}{9+1}=\frac{5}{10}$
$=\frac12$ View full question & answer→Question 134 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\pi}}\frac{\sqrt{2+\cos\text{x}}-1}{({\pi-\text{x}})^2}$
Answer$\lim\limits_{\text{x}\rightarrow{\pi}}\frac{\sqrt{2+\cos\text{x}}-1}{({\pi-\text{x}})^2}$ $=\lim\limits_{\text{x}\rightarrow{\pi}}\frac{\sqrt{2+\cos\text{x}}-1}{({\pi-\text{x}})^2}\times\frac{\sqrt{2+\cos\text{x}}+1}{{\sqrt{2+\cos\text{x}}+1}}$ $=\lim\limits_{\text{x}\rightarrow{\pi}}\frac{(2+\cos\text{x})-1}{(\pi-\text{x})^2\big(\sqrt{2+\cos\text{x}}+1\big)}$ $=\lim\limits_{\text{x}\rightarrow{\pi}}\frac{1+\cos\text{x}}{(\pi-\text{x})^2\big(\sqrt{2+\cos\text{x}}+1\big)}$ Let $\pi-\text{x}=\text{y},\text{x}\rightarrow\pi,\text{y}\rightarrow0$ $\Rightarrow\lim\limits_{\text{x}\rightarrow{\pi}}\frac{1+\cos\text{x}}{(\pi-\text{x})^2\big(\sqrt{2+\cos\text{x}}+1\big)}=\lim\limits_{\text{y}\rightarrow0}\frac{1+\cos\text{x}(\pi-\text{y})}{\text{y}^2\big(\sqrt{2+\cos(\pi-\text{y})+1}\big)}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{1-\cos\text{y}}{\text{y}^2\sqrt{2-\cos\text{y}+1}}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{\frac{2\sin^2\text{y}}{2}}{\text{y}^2\sqrt{2-\cos\text{y}}+1}$ $=2\lim\limits_{\text{y}\rightarrow0}\Bigg(\frac{\frac{\sin\text{y}}{2}}{\frac{\text{y}}{2}}\Bigg)^2\times\frac{1}{4}\frac{1}{\sqrt{2-\cos\text{y}}+1}$ $=2\times\Big(\lim\limits_{\text{y}\rightarrow0}\frac{\sin\text{y}}{2}\Big)^2\times\frac{1}{4}\frac{1}{\lim\limits_{\text{y}\rightarrow0}\sqrt{2-\cos\text{y}}+1}$ $=2\times1\times\frac{1}{4}\times\frac{1}{\sqrt{2-\cos\text{0}}+1}$ $=2\times1\times\frac{1}{4}\times\frac{1}{\sqrt{2-1}+1}$ $=2\times1\times\frac{1}{4}\times\frac{1}{{1}+1}$ $=2\times1\times\frac{1}{4}\times\frac{1}{{2}}$ $=\frac{1}{4}$
View full question & answer→Question 144 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-3\text{x}^3+2}{\text{x}^3-5\text{x}^2+3\text{x}+1}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-3\text{x}^3+2}{\text{x}^3-5\text{x}^2+3\text{x}+1}$ Dividing $\text{x}^4-3\text{x}^3+2\text{ by }\text{x}^3-5\text{x}^2+3\text{x}+1$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-3\text{x}^3+2}{\text{x}^3-5\text{x}^2+3\text{x}+1}=\lim\limits_{\text{x}\rightarrow1}\text{ x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}^2-7\text{x}}{\text{x}^3-5\text{x}^3+3\text{x}+1}$ $=\lim\limits_{\text{x}\rightarrow1}\text{x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}(\text{x}-1)}{\text{x}^3-5\text{x}^3+3\text{x}+1}$ Dividing $\text{x}^3-5\text{x}^2+3\text{x}+1\text{ by }\text{x}-1$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\text{ x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}^2-7\text{x}}{\text{x}^3-5\text{x}^3+3\text{x}+1}$ $=\lim\limits_{\text{x}\rightarrow1}\text{ x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}(\text{x}-1)}{(\text{x}-1)(\text{x}^2-4\text{x}-1)}$ $=\lim\limits_{\text{x}\rightarrow1}\text{ x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}}{\big(\text{x}^2-4\text{x}-1\big)}$ $=1+2+\frac{7}{(1-4-1)}$ $=3-\frac{7}{4}$ $=\frac{12-7}{4}$ $=\frac{5}{4}$ View full question & answer→Question 154 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\cos\text{x}-\sin\text{x}}{\big(\frac\pi4-\text{x}\big)(\cos\text{x}+\sin\text{x})}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\cos\text{x}-\sin\text{x}}{\big(\frac\pi4-\text{x}\big)(\cos\text{x}+\sin\text{x})}$$\Rightarrow\text{x}\rightarrow\frac{\pi}{4},$ then $\frac\pi4-\text{x}\rightarrow0$ let $\frac\pi4-\text{x}=\text{y}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\cos\big(\frac\pi4+\text{y}\big)-\sin\big(\frac\pi4+\text{y}\big)}{-\text{y}\big(\cos\big(\frac\pi4+\text{y}\big)+\sin\big(\frac\pi4+\text{y}\big)\big)}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\Big[\big(\cos\frac\pi4\cos\text{y}-\sin\frac\pi4\sin\text{y}\big)-\big(\sin\frac\pi4\cos\text{y}+\cos\frac\pi4\sin\text{y}\big)\Big]}{-\text{y}\big(\cos\big(\frac\pi4+\text{y}\big)+\sin\big(\frac\pi4+\text{y}\big)\big)}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\Big[\frac{\cos\text{y}}{\sqrt{2}}-\frac{\sin\text{y}}{\sqrt{2}}-\frac{\cos\text{y}}{\sqrt{2}}-\frac{\sin\text{y}}{\sqrt{2}}\Big]}{-\text{y}\big(\cos\big(\frac\pi4+\text{y}\big)+\sin\big(\frac\pi4+\text{y}\big)\big)}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\frac{-2\sin\text{y}}{\sqrt{2}}}{-\text{y}\big(\cos\big(\frac\pi4+\text{y}\big)+\sin\big(\frac\pi4+\text{y}\big)\big)}$
$=\sqrt{2}\lim\limits_{\text{y}\rightarrow{0}}\Big(\frac{\sin\text{y}}{\text{y}}\Big)\times\frac{1}{\lim\limits_{\text{y}\rightarrow{0}}\big(\cos\big(\frac\pi4+\text{y}\big)+\sin\big(\frac\pi4+\text{y}\big)\big)}$
$=\sqrt{2}\times1\times\frac{1}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}=\sqrt{2}\times\frac{1}{\frac{2}{\sqrt{2}}}$
$=\frac{\sqrt{2}\times\sqrt{2}}{2}=1$
View full question & answer→Question 164 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^7-2\text{x}^5+1}{\text{x}^3-3\text{x}^2+2}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^7-2\text{x}^5+1}{\text{x}^3-3\text{x}^2+2}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}-1)\big(\text{x}^6+\text{x}^5-\text{x}^4-\text{x}^3-\text{x}^2-\text{x}-1\big)}{(\text{x}-1)\big(\text{x}^2-2\text{x}-2\big)}$ $= \lim\limits_{\text{x}\rightarrow1}\frac{\big(\text{x}^6+\text{x}^5-\text{x}^4-\text{x}^3-\text{x}^2-\text{x}-1\big)}{\big(\text{x}^2-2\text{x}-2\big)}$ $=\frac{(1+1-1-1-1-1-1)}{(1-2-2)}$ $=\frac{-3}{-3}$ $=1$
View full question & answer→Question 174 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3+3\text{x}^2-9\text{x}-2}{\text{x}^3-\text{x}-6}$
Answer$\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3+3\text{x}^2-9\text{x}-2}{\text{x}^3-\text{x}-6}$ Dividing $\text{x}^3-3\text{x}^2+9\text{x}-2\text{ by }\text{x}^3-\text{x}-6$
$\Rightarrow\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3-3\text{x}^2-9}{\text{x}^3-\text{x}-6}=\lim\limits_{\text{x}\rightarrow2}1+\lim\limits_{\text{x}\rightarrow2}\frac{3\text{x}^2-8\text{x}+4}{\text{x}^3-\text{x}-6}$ $=1+\lim\limits_{\text{x}\rightarrow2}\frac{3\text{x}^2-8\text{x}+4}{\text{x}^3-\text{x}-6}$ $=1+\lim\limits_{\text{x}\rightarrow2}\frac{3\text{x}^2-2\text{x}-6\text{x+4}}{\text{x}^3-\text{x}-6}$ $\Rightarrow\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3-3\text{x}^2-9\text{x}-2}{\text{x}^3-\text{x}-6}-1+\lim\limits_{\text{x}\rightarrow2}\frac{(3\text{x}-2)(\text{x}-2)}{\text{x}^3-\text{x}-6}$ Dividing $\text{x}^3-\text{x}-6\text{ by}\text{ x}-2$
$\Rightarrow\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3+3\text{x}^2-9\text{x}-2}{\text{x}^3-\text{x}-6}=1+\lim\limits_{\text{x}\rightarrow2}\frac{(3\text{x}-2)(\text{x}-2)}{(\text{x}-2)(\text{x}^2+2\text{x}+3)}$ $=1+\lim\limits_{\text{x}\rightarrow2}\frac{(3\text{x}-2)}{(\text{x}^2-2\text{x}+3)}$ $=1+\frac{3\times2-2}{2^2+2\times2+3}$ $=1+\frac{4}{11}$ $=\frac{15}{11}$ View full question & answer→Question 184 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{3\sin\text{x}-\sin3\text{x}}{\text{x}^3}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{3\sin\text{x}-\sin3\text{x}}{\text{x}^3}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{3\sin\text{x}\big(3\sin\text{x}-4\sin^3\text{x}\big)}{\text{x}^3}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{4\sin^3\text{x}}{\text{x}^3}$ $=4\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}\Big)^3$ $=4\times1$ $=4$
View full question & answer→Question 194 Marks
Evaluate the following limit: If $\lim\limits_{\text{x}\rightarrow0}{\text{ kx cosec x}}=\lim\limits_{\text{x}\rightarrow0}\text{ x cosec kx,}{}$ find k.
Answer$\lim\limits_{\text{x}\rightarrow0}{\text{ kx cosec x}}=\lim\limits_{\text{x}\rightarrow0}\text{ x cosec kx,}{}$ $\lim\limits_{\text{x}\rightarrow0}\text{ kx}\frac{1}{\sin\text{x}}=\lim\limits_{\text{x}\rightarrow0}\text{x}\frac{1}{\sin\text{kx}}$ $\text{k}\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\text{x}}{\sin\text{x}}\Big)=\frac{1}{\text{k}}\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\text{kx}}{\sin\text{kx}}\Big)$ $\text{k}=\frac{1}{\text{k}}$ $\text{k}^2=1$ $\text{k}=\pm1$
View full question & answer→Question 204 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\text{x}}$$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}\big)}{\text{x}}\times\frac{\big(\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}\big)}{\big(\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(1+\text{x}^2\big)-\big(1-\text{x}^2\big)}{\text{x}\big(\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\text{x}^2}{\text{x}\big(\sqrt{1+\text{x}^2}+\sqrt{1+\text{x}^2}\big)}$
$=\frac{2\times0}{\big(\sqrt{1}+\sqrt{1}\big)}$
$=\frac{2}{2}\times0$
$=0$
View full question & answer→Question 214 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{2-\text{x}}-\sqrt{2+\text{x}}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{2-\text{x}}-\sqrt{2+\text{x}}}{\text{x}}$$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{2-\text{x}}+\sqrt{2+\text{x}}\big)\big(\sqrt{2-\text{x}}+\sqrt{2+\text{x}}\big)}{\text{x}\times\big(\sqrt{2-\text{x}}+\sqrt{2+\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(2-\text{x})-(2+\text{x})}{\text{x}\big(\sqrt{2-\text{x}}+\sqrt{2+\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{-2\text{x}}{\text{x}\big(\sqrt{2-\text{x}}+\sqrt{2+\text{x}}\big)}$
$=\frac{-2}{\sqrt{2}+\sqrt{2}}$
$=\frac{-2}{2\sqrt{2}}$
$=\frac{-1}{\sqrt{2}}$
View full question & answer→Question 224 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{5\text{x}+4\sin3\text{x}}{4\sin2\text{x}+7\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{5\text{x}+4\sin3\text{x}}{4\sin2\text{x}+7\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{5+\frac{4\sin3\text{x}}{\text{x}}}{\frac{4\sin2\text{x}}{\text{x}}+7}$ $=\frac{\lim\limits_{\text{x}\rightarrow0}5+4\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{3\text{x}}\times3}{4\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}}{2\text{x}}\times2+7}$ $=\frac{5+4\times1\times3}{4\times2+7}$ $=\frac{5+12}{8+7}$ $=\frac{17}{15}$
View full question & answer→Question 234 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{\text{cosec}^2\text{x}-2}{\cot\text{x}-1}$
Answer$\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{\text{cosec}^2\text{x}-2}{\cot\text{x}-1}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{\cot^2\text{x}+1-2}{\cot\text{x}-1}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{\cot^2\text{x}-1}{\cot\text{x}-1}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{(\cot\text{x}-1)(\cot\text{x}+1)}{\cot\text{x}-1}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}(\cot\text{x}+1)$ $=\cot\frac{\pi}{4}+1$ $=1+1$ $=2$
View full question & answer→Question 244 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}}-1}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}}-1}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+\text{x}}-1\big)}{\text{x}}\times\frac{\big(\sqrt{1+\text{x}}+1\big)}{\big(\sqrt{1+\text{x}}+1\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{(1+\text{x}-1)}{\text{x}\big(\sqrt{1+\text{x}}+1\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\text{x}\big(\sqrt{1+\text{x}}+1\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{1}{\big(\sqrt{1+\text{x}}+1\big)}$ $=\frac{1}{\sqrt{1}+1}=\frac12$
View full question & answer→Question 254 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\frac{\pi}{3}}}\frac{\sqrt{3}-\tan\text{x}}{\pi-3\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{3}}}\frac{\sqrt{3}-\tan\text{x}}{\pi-3\text{x}}$ If $\text{x}\rightarrow\frac{\pi}{3},\frac{\pi}{3}-\text{x}\rightarrow0,\pi-3\text{x}\rightarrow0$ Let $\frac\pi3-\text{x}=\text{y}$ they y → 0 $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{3}-\tan\big(\frac{\pi}{3}-\text{y}\big)}{3\big(\frac{\pi}{3}-\text{x}\big)}$ $=\lim\limits_{\text{y}\rightarrow{0}}\begin{pmatrix}\frac{\bigg(\sqrt{3}-\frac{\tan\frac\pi3-\tan\text{y}}{1+\tan\frac\pi3.\tan\text{y}}\bigg)}{3\text{y}}\end{pmatrix}$ $=\lim\limits_{\text{y}\rightarrow{0}}\begin{pmatrix}\frac{\Big(\sqrt{3}-\frac{\sqrt{3}-\tan\text{y}}{1+\sqrt{3}\tan\text{y}}\Big)}{3\text{y}}\end{pmatrix}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\big(\sqrt{3}-\tan\text{y}-\sqrt{3}+\tan\text{y}\big)}{3\big(1+\sqrt{3}\tan\text{y}\big)\text{y}}$ $=\frac{4}{3}\times\lim\limits_{\text{y}\rightarrow0}\frac{\tan\text{y}}{\text{y}}\times\frac{1}{\lim\limits_{\text{y}\rightarrow0}\Big(1+\sqrt{3}\frac{\tan\text{y}}{\text{y}}\times\text{y}\Big)}$ $=\frac{4\times1}{3}\times\frac{1}{1+0}$ $=\frac{4}{3}$
View full question & answer→Question 264 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow7}\frac{4-\sqrt{9+\text{x}}}{1-\sqrt{8-\text{x}}}$
Answer$\lim\limits_{\text{x}\rightarrow7}\frac{4-\sqrt{9+\text{x}}}{1-\sqrt{8-\text{x}}}$ $=\lim\limits_{\text{x}\rightarrow7}\frac{\big(4-\sqrt{9+\text{x}}\big)}{1-\sqrt{8-\text{x}}}\times\frac{\big(4+\sqrt{9+\text{x}}\big)}{\big(4+\sqrt{9+\text{x}}\big)}\times\frac{\big(1+\sqrt{8-\text{x}}\big)}{\big(\sqrt{1+\sqrt{8+\text{x}}}\big)}$ $=\lim\limits_{\text{x}\rightarrow7}\frac{\big((4)^2-\big(\sqrt{9+\text{x}}\big)^2\big)}{\big((1)^2-\big(\sqrt{8-\text{x}}\big)^2\big)}\times\frac{1+\sqrt{8-\text{x}}}{4+\sqrt{9+\text{x}}}$ $=\lim\limits_{\text{x}\rightarrow7}\frac{(16-9-\text{x})\times\big(1+\sqrt{8-\text{x}}\big)}{(1-8+\text{x})\times\big(4+\sqrt{9+\text{x}}\big)}$ $=\lim\limits_{\text{x}\rightarrow7}\frac{7-\text{x}}{(-7)(7-\text{x})}\frac{\big(1+\sqrt{8-\text{x}}\big)}{\big(4+\sqrt{9+\text{x}}\big)}$ $=\frac{1}{(-1)}\times\frac{(1+1)}{(4+4)}=\frac{-1}{4}$
View full question & answer→Question 274 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}}{\cos2\text{x}-\cos8\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}}{\cos2\text{x}-\cos8\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin^2\text{x}}{-2\sin\big(\frac{2\text{x}+8\text{x}}{2}\big)\sin\big(\frac{2\text{x}-8\text{x}}{2}\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\sin^2\text{x}}{\sin5\text{x}\times\sin(-3\text{x})}$ $=\frac{\lim\limits_{\text{x}\rightarrow0}\sin^2\text{x}}{-\big(\lim\limits_{\text{x}\rightarrow0}\sin5\text{x}\big)\big(-\lim\limits_{\text{x}\rightarrow0}\sin3\text{x}\big)}$ $=\frac{\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}\big)^2\times\text{x}^2}{\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin5\text{x}}{5\text{x}}\times5\text{x}\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{3\text{x}}\big)\times3\text{x}}$ $=\frac{1\times\text{x}^2}{1\times5\text{x}\times1\times3\text{x}}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$ $=\frac{\text{x}^2}{15\text{x}^2}$ $=\frac{1}{15}$
View full question & answer→Question 284 Marks
Evaluate the following limit: If $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{3}-\text{a}^3}{\text{x}-\text{a}}=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-1}{\text{x}-1},$ find all possible value of a.
AnswerIf $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{3}-\text{a}^3}{\text{x}-\text{a}}=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-1}{\text{x}-1}\ \cdots{\text{(i})}$ $\text{L.H.S}=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{3}-\text{a}^3}{\text{x}-\text{a}}$ $=3(\text{a})^{3-1}$ $=3\text{a}^{2}\ \cdots{\text{(ii})}$ $\text{R.H.S}=\lim\limits_{\text{x}\rightarrow5}\frac{\text{x}^4-1}{\text{x}-1}$ $=\lim\limits_{\text{x}\rightarrow5}\frac{\text{x}^4-1}{\text{x}-1}$ $=4(1)^{4-1}$ $=4\ \cdots{(\text{iii})}$ Substituting (ii) and (iii) in (i), $3\text{a}^8=4$ $\Rightarrow\text{a}^{2}=\frac43$ $\Rightarrow\text{a}=\pm\frac{2}{\sqrt{3}}$
View full question & answer→Question 294 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow-1}\frac{\text{x}^2-\text{x}-2}{\big(\text{x}^2+\text{x}\big)+\sin(\text{x}+1)}$
Answer$\lim\limits_{\text{x}\rightarrow-1}\frac{\text{x}^2-\text{x}-2}{\big(\text{x}^2+\text{x}\big)+\sin(\text{x}+1)}$ $=\lim\limits_{\text{x}\rightarrow-1}\frac{(\text{x}-2)(\text{x}+2)}{\text{x}(\text{x}+1)+\sin(\text{x}+1)}$ $=\lim\limits_{\text{x}\rightarrow-1}\frac{1}{\frac{\text{x}(\text{x}+1)}{(\text{x}-2)(\text{x}+1)}+\frac{\sin(\text{x}+1)}{(\text{x}-2)(\text{x}+1)}}$ $=\lim\limits_{\text{x}\rightarrow-1}\frac{1}{\frac{\text{x}}{\text{x}-2}+\frac{\sin(\text{x}+1)}{(\text{x}-2)(\text{x}+1)}}$ $=\lim\limits_{\text{x}\rightarrow-1}\frac{1}{(\text{x}-2)}\Bigg(\frac{1}{\text{x}+\frac{\sin(\text{x}+1)}{\text{x}+1}}\Bigg)$ $=\lim\limits_{\text{x}\rightarrow-1}\frac{1}{\text{x}-2}\times\frac{1}{\lim\limits_{\text{x}\rightarrow-1}(\text{x})+\lim\limits_{\text{x}\rightarrow-0}\sin\frac{\text{x}+1}{\text{x}+1}}$ $=\Big(\frac{1}{-1-2}\Big)\times\frac{1}{(-1)+1}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$ $=\frac{1}{0}$ $\Big[\because\frac10=\infty\Big]$ $=\infty$
View full question & answer→Question 304 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow\infty}\Big\{\sqrt{\text{x}+1}-\sqrt{\text{x}}\Big\}\sqrt{\text{x}+2}$
Answer$\lim\limits_{\text{x}\rightarrow\infty}\Big\{\sqrt{\text{x}+1}-\sqrt{\text{x}}\Big\}\sqrt{\text{x}+2}$ $=\lim\limits_{\text{x}\rightarrow\infty}\text{x}\Big[\sqrt{\text{x}+1}-\sqrt{\text{x}}\Big]\times\frac{\big[\sqrt{\text{x}+1}+\sqrt{\text{x}}\big]}{\big[\sqrt{\text{x}+1}+\sqrt{\text{x}}\big]}\times\frac{\sqrt{\text{x}+2}\times\sqrt{\text{x}+2}}{\sqrt{\text{x}+2}}$ $=\lim\limits_{\text{x}\rightarrow\infty}\frac{(\text{x}+1-\text{x})}{\sqrt{\text{x}+1}+\sqrt{\text{x}}}\times\frac{(\text{x}+2)}{\sqrt{\text{x}+2}}$ $=\lim\limits_{\text{x}\rightarrow\infty}\frac{1(\text{x}+2)}{\text{x}\big(\sqrt{\text{x}+1}+\sqrt{{\text{x}}}\big)\big(\sqrt{\text{x}+2}\big)}$ $=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{\text{x}\Big(1+\frac{2}{\text{x}}\Big)}{\sqrt{\text{x}}\bigg(\sqrt{1+\frac{1}{\text{x}}}+1\bigg)\bigg(\sqrt{1+\frac{2}{\text{x}}}\bigg)\sqrt{\text{x}}}$ $=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{\Big(1+\frac{2}{\text{x}}\Big)}{\bigg(\sqrt{1+\frac{1}{\text{x}}}+\sqrt{1}\bigg)\bigg(\sqrt{1+\frac{2}{\text{x}}}\bigg){}}$ $=\lim\limits_{\text{x}\rightarrow\infty}\frac{(1+0)}{(1+1)\times1}=\frac12$
View full question & answer→Question 314 Marks
Evaluate the following limit: $\lim\limits_{\text{n}\rightarrow\infty}\text{n}\sin\Big(\frac{\pi}{4\text{n}}\Big)\cos\Big(\frac{\pi}{4\text{n}}\Big)$
Answer$\lim\limits_{\text{n}\rightarrow\infty}\text{n}\sin\Big(\frac{\pi}{4\text{n}}\Big)\cos\Big(\frac{\pi}{4\text{n}}\Big)$ $=\lim\limits_{\text{n}\rightarrow\infty}2\Big(\text{n}\sin\frac{\pi}{4\text{n}}\cos\frac{\pi}{4\text{n}}\Big)\times\frac12$ $=\lim\limits_{\text{n}\rightarrow\infty}\text{n}\times\sin\frac{\pi}{2\text{n}}\times\frac12$ $\text{n}\rightarrow\infty,$ then $\frac{1}{\text{n}}\rightarrow0, $ let $\frac{1}{\text{n}}=\text{y}$ $=\frac12\lim\limits_{\frac{1}{\text{n}}\rightarrow\infty}\frac{1}{\text{y}}\sin\Big(\frac{\pi}{2}\Big)\Big(\frac{1}{\text{n}}\Big)$ $=\frac12\lim\limits_{{\text{y}}\rightarrow\infty}\frac{\sin\big(\frac\pi2\big)\text{y}}{\text{y}}$ $=\frac12\Bigg(\lim\limits_{{\text{y}}\rightarrow\infty}\frac{\sin\big(\frac{\pi\text{y}}2\big)}{\frac{\pi\text{y}}{2}}\Bigg)\times\frac\pi2$ $=\frac12\times1\times\frac\pi2$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$ $=\frac\pi4$
View full question & answer→Question 324 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{27}}\frac{\Big(\text{x}^\frac{1}{3}+3\Big)\Big(\text{x}^{\frac{1}{3}}-3\Big)}{\text{x}-27}$
Answer$\lim\limits_{\text{x}\rightarrow{27}}\frac{\Big(\text{x}^\frac{1}{3}+3\Big)\Big(\text{x}^{\frac{1}{3}}-3\Big)}{\text{x}-27}$ $=\lim\limits_{\text{x}\rightarrow{27}}\frac{\Big(\text{x}^\frac{1}{3}+9\Big)}{\text{x}-27}$ $=\lim\limits_{\text{x}\rightarrow{27}}\frac{\text{x}^\frac{2}{3}-27^{\frac{2}{3}}}{\text{x}-27}$ Applying formula $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{n}}-\text{a}^\text{n}}{\text{x}-\text{a}}=\text{na}^{\text{n}-1}$ $=\frac23(27)^{\frac23-1}$ $=\frac23(27)^{\frac{-1}{3}}$ $=\frac23\times\frac{1}{(27)^{\frac{1}{3}}}$ $=\frac{2}{3}\times\frac{1}{3}$ $=\frac29$
View full question & answer→Question 334 Marks
Evaluate the following limit: $\lim\limits_{\text{n}\rightarrow\infty}{\Big(\frac{1}{\text{n}^2}+\frac{2}{\text{n}^2}+\frac{3}{\text{n}^2}+\ \cdots+\frac{\text{n}-1}{\text{n}^2}}{}\Big)$
Answer$\lim\limits_{\text{n}\rightarrow\infty}{\Big(\frac{1}{\text{n}^2}+\frac{2}{\text{n}^2}+\frac{3}{\text{n}^2}+\ \cdots+\frac{\text{n}-1}{\text{n}^2}}{}\Big)$$=\lim\limits_{\text{n}\rightarrow\infty}\Big(\frac{1+2+3+\ \cdots+(\text{n}-1)}{{n}^2}\Big)$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{\big(\text{n}+1\big)(\text{n})}{2\times\text{n}^2}$ $\Big[1+2+3+\ \cdots+({\text{n}}-1)=\frac{({\text{n}}-1)({\text{n}})}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{\text{n}^2-\text{n}}{2\text{n}^2}$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$
$=\lim\limits_{\text{n}\rightarrow{\infty}}\frac{1-\frac{1}{\text{n}}}{2}$
$=\frac{1-0}{2}=\frac12$
$=\frac{1}{2}$
View full question & answer→Question 344 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\frac{\pi}{6}}}\frac{\cot^2\text{x}-3}{\text{cosec x}-2}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{6}}}\frac{\cot^2\text{x}-3}{\text{cosec x}-2}$ $=\lim\limits_{\text{x}\rightarrow{\frac{\pi}{6}}}\frac{\big(\text{cosec}^2\text{x}-1\big)-3}{\text{cosec x}-2}$ $=\lim\limits_{\text{x}\rightarrow{\frac{\pi}{6}}}\frac{\big(\text{cosec}^2\text{x}-4\big)}{\text{cosec x}-2}$ $=\lim\limits_{\text{x}\rightarrow{\frac{\pi}{6}}}\frac{\big(\text{cosec }\text{x}-2\big)\big(\text{cosec }\text{x}+2\big)}{\text{cosec x}-2}$ $=\lim\limits_{\text{x}\rightarrow{\frac{\pi}{6}}}\text{cosec }\text{x}+2$ $=\text{cosec}\frac{\pi}{6}+2$ $=2+2$ $=4$
View full question & answer→Question 354 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow\infty}\text{x}\Big\{\sqrt{\text{x}^2+1}-\sqrt{\text{x}^2-1}\Big\}$
Answer$\lim\limits_{\text{x}\rightarrow\infty}\text{x}\Big\{\sqrt{\text{x}^2+1}-\sqrt{\text{x}^2-1}\Big\}$ $=\lim\limits_{\text{x}\rightarrow\infty}\text{x}\Big[\sqrt{\text{x}^2+1}-\sqrt{\text{x}^2-1}\Big]\times\frac{\big(\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}\big)}{\big(\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2+1}\big)}$ $=\lim\limits_{\text{x}\rightarrow\infty}\text{x}\frac{\text{x}\big(\text{x}^2+1-\text{x}^2+1\big)}{\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}}$ $=\lim\limits_{\text{x}\rightarrow\infty}\frac{\text{x}(2)}{\text{x}\Big(\sqrt{1+\frac{1}{\text{x}^2}}+\sqrt{1-\frac{1}{\text{x}^2}}\Big)}$ $=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{2}{\sqrt{1+\frac{1}{\text{x}^2}}+\sqrt{1-\frac{1}{\text{x}^2}}}$ $=\frac{2}{2}=1$ $=1$
View full question & answer→Question 364 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{{\pi}}}\frac{1+\cos\text{x}}{\tan^2\text{ x}}$
Answer$\lim\limits_{\text{x}\rightarrow{{\pi}}}\frac{1+\cos\text{x}}{\tan^2\text{ x}}$ $\Rightarrow\text{x}\rightarrow{\pi},\text{x}-{\pi}\rightarrow0,$ let $\text{y}=\text{x}-\pi$ $\Rightarrow\lim\limits_{\text{x}-{\pi}\rightarrow{{0}}}\frac{1+\cos\text{x}}{\tan^2\text{x}}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{1+\cos(\pi+\text{y})}{\tan^2(\pi+\text{y})}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{1-\cos\text{y}}{\tan^2\text{y}}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\frac{\text{y}}{2}}{\tan^2\text{y}}$ $=2\Bigg(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\frac{\text{y}}{2}}{\frac{\text{y}}{2}}\Bigg)^2\times\frac{\text{y}^2}{4}\times\frac{1}{\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\tan\text{y}}{\text{y}}\Big)^2\times\text{y}^2}$ $=2\times1\times\frac{\text{y}^2}{4}\times\frac{1}{1\times\text{y}^2}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1,\lim\limits_{\theta\rightarrow0}\frac{\tan\theta}{\theta}=1\Big]$ $=\frac{1}{2}$
View full question & answer→Question 374 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\sqrt{1+\text{x}}-\sqrt{1-\text{x}}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\sqrt{1+\text{x}}-\sqrt{1-\text{x}}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\big(\sqrt{1+\text{x}}-\sqrt{1-\text{x}}\big)}\times\frac{\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}{\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}{\big(\sqrt{1+\text{x}}\big)^2-\big(\sqrt{1-\text{x}}\big)^2}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}{1+\text{x}-1+\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}{2\text{x}}$ $=\frac12\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\sqrt{1+\text{x}}+\sqrt{1-\text{x}}}{\text{x}}\Big)\text{x}$ $=\frac{1}{2}\lim\limits_{\text{x}\rightarrow0}{\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}$ $=\frac{1}{2}\big(\sqrt{1}+\sqrt{1}\big)$ $=\frac{1}{2}(1+1)=\frac22$ $=1$
View full question & answer→Question 384 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow\pi}\frac{\sqrt{5+\cos\text{x}-2}}{(\pi-\text{x})^2}$
Answer$\lim\limits_{\text{x}\rightarrow\pi}\frac{\sqrt{5+\cos\text{x}-2}}{(\pi-\text{x})^2}$ $⇒ \text{x} → \pi,$ then $\pi-\text{x}\rightarrow0,$ let $\pi-\text{x}=\text{y}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{5+\cos(\pi-\text{y})}-2}{\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{5-\cos\text{y}}-2}{\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{5-\cos\text{y}}-2}{\text{y}^2}\times\frac{\big(\sqrt{5-\cos\text{y}}+2\big)}{\big(\sqrt{5-\cos\text{y}}+2\big)}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{(5-\cos\text{y}-4)}{\text{y}^2\big(\sqrt{5-\cos\text{y}}+2\big)}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\frac{\text{y}}{2}}{\text{y}^2\big(\sqrt{5-\cos\text{y}}+2\big)}$ $=2\times\frac14\times\frac{1}{\sqrt{4}+2}=2\times\frac14\times\frac14$ $=\frac18$
View full question & answer→Question 394 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}-2}{\sqrt{\text{x}}-\sqrt{2}}$
Answer$\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}-2}{\sqrt{\text{x}}-\sqrt{2}}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}-2)\big(\sqrt{\text{x}}+\sqrt{2}\big)}{\big(\sqrt{\text{x}}-\sqrt{2}\big)\big(\sqrt{\text{x}}+\sqrt{2}\big)}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}-2)\big(\sqrt{\text{x}}+\sqrt{2}\big)}{(\text{x}-2)}$ $=\sqrt{2}+\sqrt{2}$ $=2\sqrt{2}$
View full question & answer→Question 404 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\tan\text{x}}{1-\cos2\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\tan\text{x}}{1-\cos2\text{x}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}\tan\text{x}}{2\sin^2\text{x}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{\frac{\tan\text{x}}{\text{x}}}{\frac{2\sin^2\text{x}}{\text{x}^2}}$ $=\frac{\lim\limits_{\text{x} \rightarrow0}\frac{\tan\text{x}}{\text{x}}}{2\lim\limits_{\text{x} \rightarrow0}\big(\frac{\sin\text{x}}{\text{x}}\big)^2}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1,\lim\limits_{\theta\rightarrow0}\frac{\tan\theta}{\theta}=1\Big]$ $=\frac{1}{2\times1}$ $=\frac{1}{2}$
View full question & answer→Question 414 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\cos\text{ax}-\cos\text{bx}}{\cos\text{cx}-1}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\cos\text{ax}-\cos\text{bx}}{\cos\text{cx}-1}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{1-2\sin^2\big(\frac{\text{ax}}{2}\big)-1+2\sin^2\big(\frac{\text{bx}}{2}\big)}{1-2\sin^2\big(\frac{\text{cx}}{2}\big)-1}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{-2\sin^2\big(\frac{\text{ax}}{2}\big)+2\sin^2\big(\frac{\text{bx}}{2}\big)}{-2\sin^2\big(\frac{\text{cx}}{2}\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{-\sin^2\big(\frac{\text{ax}}{2\text{ax}}\big)4\text{a}^2\text{x}^2+\sin\big(\frac{\text{bx}}{2}\big)4\text{b}^2\text{x}^2}{-\sin^2\big(\frac{\text{cx}}{2}\big)4\text{c}^2\text{x}^2}$ $=\frac{-\text{a}^2+\text{b}^2}{-\text{c}^2}$ $=\frac{\text{a}^2-\text{b}^2}{\text{c}^2}$
View full question & answer→Question 424 Marks
Evaluate the following limit: If $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{9}-\text{a}^9}{\text{x}-\text{a}}=\lim\limits_{\text{x}\rightarrow5}(4+\text{x}),$ find all possible value of a.
AnswerIf $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{9}-\text{a}^9}{\text{x}-\text{a}}=\lim\limits_{\text{x}\rightarrow5}(4+\text{x})\ \cdots{\text{(i})}$ $\text{L.H.S}=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{9}-\text{a}^9}{\text{x}-\text{a}}$ $=9(\text{a})^{9-1}$ $=9\text{a}^{8}\ \cdots{\text{(ii})}$ $\text{R.H.S}=\lim\limits_{\text{x}\rightarrow5}(4+\text{x})$ $=4+5=9\ \cdots{\text{(iii})}$ Substituting (ii) and (iii) in (i) $9\text{a}^8=9$ $\Rightarrow\text{a}^{8}=1$ $\Rightarrow\text{a}^4=1$ $\text{a}^2=1$ $\Rightarrow\text{a} = 1\text{ and a} = -1$
View full question & answer→Question 434 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\text{x}^2}{\sin2\pi\text{ x}}$
Answer$\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\text{x}^2}{\sin2\pi\text{ x}}$ ⇒ x → 1, then x - 1 →0, let x - 1 = y $=\lim\limits_{(\text{x}-1)\rightarrow{0}}\frac{(1-\text{x})(1+\text{x})}{\sin2\pi\text{ x}}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{-\text{y}(1+\text{y}+1)}{\sin2\pi(\text{y}+1)}$ $=-\lim\limits_{\text{y}\rightarrow{0}}\frac{\text{y}(\text{y}+2)}{\sin(2\pi\text {y}+2\pi)}$ $=-\lim\limits_{\text{y}\rightarrow{0}}\frac{\text{y}(\text{y}+2)}{\sin2\pi\text {y}}$ $=-\lim\limits_{\text{y}\rightarrow{0}}(\text{y}+2)\times\frac{\text{y}}{\Big(\lim\limits_{\text{y}\rightarrow{0}}\sin\frac{2\pi\text {y}}{\text {y}\times2\pi}\Big)\times2\pi\text {y}}$ $=-2\times\frac{1}{1\times2\pi}$ $=-\frac{1}{\pi}$
View full question & answer→Question 444 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}+\text{x}^2}-\sqrt{\text{x}+1}}{2\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}+\text{x}^2}-\sqrt{\text{x}+1}}{2\text{x}^2}$ $\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+\text{x}+\text{x}^2}-\sqrt{\text{x}+1}\big)\times\big(\sqrt{1+\text{x}+\text{x}^2}+\sqrt{\text{x}+1}\big)}{2\text{x}^2\big(\sqrt{1+\text{x}+\text{x}^2}+\sqrt{\text{x}+1}\big)}$ $\lim\limits_{\text{x}\rightarrow0}\frac{\big(1+\text{x}+\text{x}^2\big)-(\text{x}+1)}{2\text{x}^2\big(\sqrt{1+\text{x}+\text{x}^2}+\sqrt{\text{x}+1}\big)}$ $\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2}{2\text{x}^2\big(\sqrt{1+\text{x}+\text{x}^2}+\sqrt{\text{x}+1}\big)}$ $=\frac{1}{2\big(\sqrt{1}+\sqrt{1}\big)}$ $=\frac{1}{2\times2}$ $=\frac14$
View full question & answer→Question 454 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{1-\sin2\text{x}}{1+\cos4\text{ x}}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{1-\sin2\text{x}}{1+\cos4\text{ x}}$ $\Rightarrow\text{x}\rightarrow\frac{\pi}{4},\text{x}-\frac{\pi}{4}\rightarrow0,$ let $\text{x}-\frac{\pi}{4}=\text{y}$ $\Rightarrow\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{1-\sin2\text{x}}{1+\cos4\text{x}}=\lim\limits_{\text{x}-\frac{\pi}{4}\rightarrow{0}}\frac{\Big(1-\sin2\big(\text{y}+\frac{\pi}{4}\big)\Big)}{1+\cos4\big(\text{y}+\frac{\pi}{4}\big)}$ $=\lim\limits_{\text{x}-\frac{\pi}{4}\rightarrow{0}}\Bigg(\frac{1-\sin\big(\frac\pi2+2\text{y}\big)}{1+\cos(\pi+4\text{y})}\Bigg)$ $=\lim\limits_{\text{x}-\frac\pi4\rightarrow{0}}\frac{1-\cos2\text{y}}{1-\cos4\text{y}}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\text{y}}{2\sin^22\text{y}}$ $=\frac{\lim\limits_{\text{y}\rightarrow{0}}\sin^2\text{y}}{\lim\limits_{\text{y}\rightarrow{0}}\sin^22\text{y}}$ $=\frac{\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\text{y}}{\text{y}}\Big)^2\times\text{y}^2}{\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin2\text{y}}{2\text{y}}\Big)^2\times4\text{y}^2}$ $=\frac{1\times\text{y}^2}{1\times4\text{y}^2}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$ $=\frac14$
View full question & answer→Question 464 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}+\tan^2\text{x}}{\text{x}\sin\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}+\tan^2\text{x}}{\text{x}\sin\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin^2\text{x}+\tan^2\text{x}}{\text{x}\sin\text{x}}$ $=\frac{2\lim\limits_{\text{x}\rightarrow0}\sin^2\text{x}+\lim\limits_{\text{x}\rightarrow0}\tan^2\text{x}}{\lim\limits_{\text{x}\rightarrow0}\text{x}\sin\text{x}}$ $=\frac{\Big(2\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}\big)^2\times\text{x}^2\Big)+\big(\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}\big)^2\times\text{x}^2}{\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}\big)\times\text{x}^2}$ $=\frac{\big(2\times1\times\text{x}^2\big)+\big(1\times\text{x}^2\big)}{\big(1\times\text{x}^2\big)}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\text{ and }\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$ $=\frac{3\text{x}^2}{\text{x}^2}$ $=3$
View full question & answer→Question 474 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{{\pi}}}\frac{\sqrt{2+\cos\text{x}-1}}{(\pi-\text{x})^2}$
Answer$\lim\limits_{\text{x}\rightarrow{{\pi}}}\frac{\sqrt{2+\cos\text{x}-1}}{(\pi-\text{x})^2}$ $\text{x}\rightarrow{\pi},$ then $\text{x}-{\pi}\rightarrow0,$ let $\text{x}-\pi=\text{y}$ $\Rightarrow\lim\limits_{{\text{x}\rightarrow{\pi}}}\frac{\big(\sqrt{2+\cos\text{x}}-1\big)}{(\pi-\text{x})^2}=\lim\limits_{{\text{x}-{\pi}}\rightarrow0}\frac{\sqrt{2+\cos(\text{x})}-1}{(-1)^2(\text{x}-\pi)^2}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{\sqrt{2+\cos(\pi+\text{y})}-1}{\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{\sqrt{2-\cos\text{y}}-1}{\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{\big(\sqrt{2-\cos\text{y}}-1\big)\big(\sqrt{2-\cos\text{y}}+1\big)}{\text{y}^2\big(\sqrt{2\cos\text{y}}+1\big)}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{(2\cos\text{y}-1)}{\big(\sqrt{2-\cos\text{y}+1}\big)\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{(1-\cos\text{y})}{\big(\sqrt{2-\cos\text{y}}+1\big)\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{2\sin^2\frac{\text{y}}{2}}{\text{y}^2\big(\sqrt{2-\cos\text{y}}+1\big)}$ $=2\lim\limits_{\text{y}\rightarrow0}\Bigg(\frac{\frac{\sin\text{y}}{2}}{\frac{\text{y}}{2}}\Bigg)^2\times\frac14\times\frac{1}{\lim\limits_{\text{y}\rightarrow0}\sqrt{2-\cos0+1}}$ $=2\times\frac14\times\frac{1}{\sqrt{2}-1+1}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$ $=\frac{1}{4}$
View full question & answer→Question 484 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^2-\sqrt{\text{x}}}{\sqrt{\text{x}}-1}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^2-\sqrt{\text{x}}}{\sqrt{\text{x}}-1}$$=\lim\limits_{\text{x}\rightarrow1}\frac{\big(\text{x}^2-\sqrt{\text{x}}\big)\big(\text{x}^2+\sqrt{\text{x}}\big)}{\big(\sqrt{\text{x}}-1\big)\big(\text{x}^2+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-\text{x}}{\big(\sqrt{\text{x}-1}\big)\big({\text{x}^2}+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}\big(\text{x}^3-1\big)}{\big(\sqrt{\text{x}-1}\big)\big(\text{x}^2+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}(\text{x}-1)\big(\text{x}^2+1+\text{x}\big)}{\big(\sqrt{\text{x}-1}\big)\big(\text{x}^2+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}\big(\sqrt{\text{x}+1}\big)\big({\text{x}^2+1+\text{x}}\big)}{\big(\text{x}^2+\sqrt{\text{x}}\big)}$
$=\frac{1(1+1)(1+1+1)}{1+1}$
$=\frac62$
$=3$
View full question & answer→Question 494 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow\frac{\pi}{6}}\frac{\text{cosec}^2\text{ x}-3}{\text{cosec}\text{ x}-2}$
Answer$\lim\limits_{\text{x}\rightarrow\frac{\pi}{6}}\frac{\text{cosec}^2\text{x}-3}{\text{cosec }\text{x}-2}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{6}}\frac{\text{cosec}^2\text{x}-4}{\text{cosec }\text{x}-2}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{6}}\frac{(\text{cosec }\text{x}-2)(\text{cosec x}+2)}{(\text{cosec }\text{x}-2)}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{6}}(\text{cosec }\text{x}+2)$ $=\text{cosec}\frac{\pi}{6}+2$ $=2+2=4$
View full question & answer→Question 504 Marks
Evaluate the following limit: $\lim\limits_{\text{n}\rightarrow\infty}\frac{1^3+2^3+\ \dots+\text{n}^3}{(\text{n}-1)^4}$
Answer$\lim\limits_{\text{n}\rightarrow\infty}\frac{1^3+2^3+3^3\ \dots+\text{n}^3}{(\text{n}-1)^4}$ $=\lim\limits_{\text{n}\rightarrow\infty}\frac{\Big[\frac12(\text{n})(\text{n}+1)\Big]^2}{(\text{n}-1)^4}$ $\bigg[1^3+2^3+3^3+\ \cdots+\text{n}^3=\Big(\frac12\text{n}(\text{n}+1)\Big)^2\bigg]$ $=\lim\limits_{\text{n}\rightarrow\infty}\Bigg(\frac{\frac14\text{n}^2\big(\text{n}^2+1+2\text{n}\big)}{(\text{n}-1)^4}\Bigg)$ $=\frac14\lim\limits_{\text{n}\rightarrow\infty}\bigg(\frac{\text{n}^4+\text{n}^2+2\text{n}^3}{(\text{n}-1)^2(\text{n}-1)^2}\bigg)$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$ $=\frac{1}{4}\lim\limits_{\text{n}\rightarrow{\infty}}\Big(\frac{\text{n}^4+\text{n}^2+2\text{n}^3}{\text{n}^4+\text{n}^2-2\text{n}^3+\text{n}^2+1-2\text{n}-2\text{n}^3-2\text{n}+4\text{n}^2}\Big)$ $=\frac{1}{4}\lim\limits_{\text{n}\rightarrow{\infty}}\frac{\Big(1+\frac{1}{\text{n}^2}+\frac{2}{\text{n}}\Big)}{\Big(1+\frac{1}{\text{n}^2}+\frac{2}{\text{n}}+\frac{1}{\text{n}^2}+\frac{1}{\text{n}^4}-\frac{2}{\text{n}^3}-\frac{2}{\text{n}}-\frac{2}{\text{n}^3}+\frac{4}{\text{n}^2}\Big)}$ $=\frac{1}{4}\Big(\frac14\Big)$ $=\frac14$
View full question & answer→