Question 514 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{cosec x}-\cot\text{x}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{cosec x}-\cot\text{x}}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}\Big)\times\frac{1}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{1}{\sin\text{x}}\Big(\frac{1-\cos\text{x}}{\text{x}}\Big)\Big)$ $=\lim\limits_{\text{x}\rightarrow0}\bigg(\frac{1}{\sin\text{x}}\bigg(\frac{2\sin^2\frac{\text{x}}{2}}{\text{x}}\bigg)\bigg)$ $=2\lim\limits_{\text{x}\rightarrow0}\Bigg(\frac{1}{\frac{\sin\text{x}}{\text{x}}}\times\text{x}\bigg(\frac{\sin\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\bigg)\times\frac{\text{x}}{4}\Bigg)$ $=2\bigg(\lim\limits_{\text{x}\rightarrow0}\frac{1}{\frac{\sin\text{x}}{\text{x}}}\bigg)\times\frac{1}{\text{x}}\times\bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\bigg)\times\frac{\text{x}}{4}$ $=2\times\frac{1}{\text{x}}\times\frac{\text{x}}{4}$ $=\frac{1}{2}$
View full question & answer→Question 524 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\cos\text{ax}-\cos\text{bx}}{\cos\text{cx}-\cos\text{dx}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\cos\text{ax}-\cos\text{bx}}{\cos\text{cx}-\cos\text{dx}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{{\Big(-2\sin\big(\frac{\text{a}+\text{b}}{2}\big)\times\sin\big(\frac{\text{a}+\text{b}}{2}\big)\text{x}\Big)}}{-2\sin\big(\frac{\text{c}+\text{d}}{2}\big)\times\big(\frac{\text{c}-\text{d}}{2}\big)\text{x}}$ $=\frac{\lim\limits_{\text{x}\rightarrow0}{\sin\big(\frac{\text{a}+\text{b}}{2}\big)\text{x}\sin\big(\frac{\text{a}+\text{b}}{2}\big)\text{x}}}{\lim\limits_{\text{x}\rightarrow0}{\sin\big(\frac{\text{c}+\text{d}}{2}\big)\text{x}\sin\big(\frac{\text{c}+\text{d}}{2}\big)\text{x}}}$ $=\frac{\Bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\big(\frac{\text{a}+\text{b}}{2}\big)\text{x}}{\big(\frac{\text{a}+\text{b}}{2}\big)\text{x}}\times\big(\frac{\text{a}+\text{b}}{2}\big)\text{x}\Bigg)\Bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\big(\frac{\text{a}-\text{b}}{2}\big)\text{x}}{\big(\frac{\text{a}-\text{b}}{2}\big)\text{x}}\times\big(\frac{\text{a}-\text{b}}{2}\big)\text{x}\Bigg)}{\Bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\big(\frac{\text{c}+\text{d}}{2}\big)\text{x}}{\big(\frac{\text{c}+\text{d}}{2}\big)\text{x}}\times\big(\frac{\text{c}+\text{d}}{2}\big)\text{x}\Bigg)\Bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\big(\frac{\text{c}-\text{d}}{2}\big)\text{x}}{\big(\frac{\text{c}-\text{d}}{2}\big)\text{x}}\times\big(\frac{\text{c}-\text{d}}{2}\big)\text{x}\Bigg)}$ $=\frac{(\text{a}+\text{b})(\text{a}-\text{b})}{(\text{c}+\text{d})(\text{c}-\text{d})}$ $\Big[\because\ \lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$ $=\frac{\text{a}^2-\text{b}^2}{\text{c}^2-\text{d}^2}$
View full question & answer→Question 534 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{3}}-\text{a}^{\frac{2}{3}}}{\text{x}^{\frac{3}{4}}-\text{a}^{\frac{3}{4}}}$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{3}}-\text{a}^{\frac{2}{3}}}{\text{x}^{\frac{3}{4}}-\text{a}^{\frac{3}{4}}}$$=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\frac{\text{x}^{\frac{2}{3}}-\text{a}^{\frac{2}{3}}}{\text{x}-\text{a}}}{\frac{\text{x}^\frac{3}{4}-\text{a}^\frac{3}{4}}{\text{x}-\text{a}}}$ [Dividing numerator and denominator by x - a]
$=\frac{\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{3}}-\text{a}^{\frac{2}{3}}}{\text{x}-\text{a}}}{\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^\frac{3}{4}-\text{a}^\frac{3}{4}}{\text{x}-\text{a}}}$
Applying the 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}$ in numerator and $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{m}}-\text{a}^{\text{m}}}{\text{x}-\text{a}}=\text{ma}^{\text{m}-1}$ in denominator respectively
Here, $\text{n}=\frac{2}{3},\text{m}=\frac{3}{4}$
$\Rightarrow\frac{\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{3}}-\text{a}^{\frac{2}{3}}}{\text{x}-\text{a}}}{\lim\limits_{\text{x}\rightarrow{\text{x}}}\frac{\text{x}^{\frac{3}{4}}-\text{a}^{\frac{3}{4}}}{\text{x}-\text{a}}}=\frac{\frac{2}{3}(\text{a})^{\frac{2}{3}-1}}{\frac{3}{4}(\text{a})^{\frac{3}{4}-1}}$
$=\frac{8}{9}\text{a}^{\frac{-1}{3}+\frac{1}{4}}$
$=\frac89\text{a}^{\frac{-1}{12}}$
View full question & answer→Question 544 Marks
Find $\lim\limits_{\text{x}\rightarrow-\frac{5}{2}}{[\text{x}]}.$
Answer$\lim\limits_{\text{x}\rightarrow-\frac{5}{2}}{[\text{x}]}$ $\lim\limits_{\text{x}\rightarrow\frac{5}{2}}{[\text{x}]}=\Big[\frac{5}{2}\Big],$ $=[2.5]=2$ [By definition of geatest integer function] $\Rightarrow\ \lim\limits_{\text{x}\rightarrow\frac{5}{2}}{[\text{x}]}=2$
View full question & answer→Question 554 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{a}\sin\text{x}-\text{x}\sin\text{a}}{\text{ax}^2-\text{xa}^2}$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{a}\sin\text{x}-\text{x}\sin\text{a}}{\text{ax}^2-\text{xa}^2}$ $=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{(\text{a}\sin\text{x}-\text{x}\sin\text{a})}{\text{ax}(\text{x}-\text{a})}$ If t = x - a Then, as x → a, t → 0 $\therefore\ \lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{(\text{a}\sin\text{x}-\text{x}\sin\text{a})}{\text{ax}(\text{x}-\text{a})}$ $=\lim\limits_{\text{t}\rightarrow{0}}\frac{\big(\text{a}\sin(\text{t}+\text{a})-(\text{t}+\text{a})\sin\text{a}\big)}{\text{a}(\text{t}+\text{a})\text{t}}$ $=\lim\limits_{\text{t}\rightarrow{0}}\frac{\text{a}\sin\text{t}\cos\text{a}+\text{a}\sin\text{a}\cot\text{t}-\text{t}\sin\text{a}-\text{a}\sin\text{a}}{\text{a}(\text{t}+\text{a})\text{a}}$ $=\lim\limits_{\text{t}\rightarrow{0}}\frac{\text{a}\sin\text{t}\cos\text{a}+\text{a}\sin\text{a}(\cos\text{t}-1)-\text{t}\sin\text{a}}{\text{a}(\text{t}+\text{a})\text{t}}$ $=\lim\limits_{\text{t}\rightarrow0}\frac{\text{a}\sin\cos\text{a}+\text{a}\sin\text{a}\Big(2\sin^2\big(\frac{\text{t}}{2}\big)\Big)-\text{t}\sin\text{a}}{\text{a}(\text{t}+\text{a})\text{t}}$ $=\lim\limits_{\text{t}\rightarrow0}\frac{\text{a}\sin\text{t}\cos\text{a}}{\text{a}(\text{t}+\text{a})\text{t}}+\lim\limits_{\text{t}\rightarrow0}\frac{\text{a}\sin\text{a}\Big(2\sin^2\big(\frac{\text{t}}{2}\big)\Big)}{\text{a}(\text{t}+\text{a})\text{t}}-\lim\limits_{\text{t}\rightarrow0}\frac{\text{t}\sin\text{a}}{\text{a}(\text{t}+\text{a})\text{t}}$ $=\frac{\text{a}\cos\text{a}}{\text{a}^2}+0-\frac{\sin\text{a}}{\text{a}^2}$ $=\frac{\text{a}\cos\text{a}-\sin\text{a}}{\text{a}^2}$
View full question & answer→Question 564 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{3+\text{x}}-\sqrt{5-\text{x}}}{\text{x}^2-1}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{3+\text{x}}-\sqrt{5-\text{x}}}{\text{x}^2-1}$$=\lim\limits_{\text{x}\rightarrow1}\frac{\big(\sqrt{3+\text{x}}-\sqrt{5-\text{x}}\big)}{\big(\text{x}^2-1\big)}\times\frac{\big(\sqrt{3+\text{x}}+\sqrt{5-\text{x}}\big)}{\big(\sqrt{3+\text{x}}+\sqrt{5-\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{(3+\text{x})-(5-\text{x})}{(\text{x}-1)(\text{x}+1)\big(\sqrt{3+\text{x}}+\sqrt{5+\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{-2+2\text{x}}{(\text{x}-1)(\text{x}+1)\big(\sqrt{3+\text{x}}+\sqrt{5+\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{2}{(\text{x}+1)\big(\sqrt{3+\text{x}}+\sqrt{5+\text{x}}\big)}$
$=\frac{2}{(1+1)\big(\sqrt{3+1}+\sqrt{5-1}\big)}=\frac{2}{(2)(2+2)}$
$=\frac14$
View full question & answer→Question 574 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}^\circ-\sin2\text{x}^\circ}{\text{x}^3}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}^\circ-\sin2\text{x}^\circ}{\text{x}^3}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin\frac{\pi\text{x}}{180}-\sin\frac{2\pi\text{x}}{180}}{\text{x}^3}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin\frac{\pi\text{x}}{180}-2\sin\frac{\pi\text{x}}{180}\cos\frac{\pi\text{x}}{180}}{\text{x}^3}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin\frac{\pi\text{x}}{180}\big(2\sin^2\frac{\pi\text{x}}{360}\big)}{\text{x}^3}$ $=4\bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\frac{\pi\text{x}}{180}}{\text{x}}\bigg)\times\bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\frac{\pi\text{x}}{360}}{\text{x}}\bigg)\times\bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\frac{\pi\text{x}}{360}}{\text{x}}\bigg)$ $=4\Bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\frac{\pi\text{x}}{180}}{\frac{\pi\text{x}}{180}}\times\frac{\pi}{180}\Bigg)\times\Bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\frac{\pi\text{x}}{360}}{\frac{\pi\text{x}}{360}}\times\frac{\pi}{360}\Bigg)\times\Bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\frac{\pi\text{x}}{360}}{\frac{\pi\text{x}}{360}}\times\frac{\pi}{360}\Bigg)$ $=4\times\frac{\pi}{180}\times\frac{\pi}{360}\times\frac{\pi}{360}$ $=\Big(\frac{\pi}{180}\Big)^3$
View full question & answer→Question 584 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3\text{x}^2+3\text{x}-6}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3\text{x}^2+3\text{x}-6}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3\big[\text{x}^2+\text{x}-2\big]}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3\big[\text{x}^2+2\text{x}-\text{x}-2\big]}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3(\text{x}+2)(\text{x}-1)}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3(\text{x}+2)\big(\sqrt{\text{x}-1}\big)\big(\sqrt{\text{x}+1}\big)}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)}{3(\text{x}+2)\big(\sqrt{\text{x}}+1\big)}$ $=\frac{(2-3)}{3(1+2)(1+1)}$ $=\frac{-1}{3\times3\times2}$ $=-\frac{1}{18}$
View full question & answer→Question 594 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow2}\frac{\sqrt{\text{x}^2+1}-\sqrt{5}}{\text{x}-2}$
Answer$\lim\limits_{\text{x}\rightarrow2}\frac{\sqrt{\text{x}^2+1}-\sqrt{5}}{\text{x}-2}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{\sqrt{\text{x}^2+1}-\sqrt{5}}{\text{x}-2}\times\frac{\big(\sqrt{\text{x}^2+1}+\sqrt{5}\big)}{\big(\sqrt{\text{x}^2+1}+\sqrt{5}\big)}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{\big({\text{x}^2+1}-{5}\big)}{(\text{x}-2)\big(\sqrt{\text{x}^2+1}+\sqrt{5}\big)}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}+2)(\text{x}-2)}{(\text{x}-1)\big(\sqrt{\text{x}^2+1}+\sqrt{5}\big)}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}+2)}{\big(\sqrt{\text{x}^2+1}+\sqrt{5}\big)}$ $=\frac{(2+2)}{\sqrt{4+1}+\sqrt{5}}$ $=\frac{4}{2\sqrt{5}}=\frac{2}{\sqrt{5}}$
View full question & answer→Question 604 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\sqrt{2\sin\text{x}}-1}{\big(\frac{\pi}{2}-\text{x}\big)}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\sqrt{2\sin\text{x}}-1}{\big(\frac{\pi}{2}-\text{x}\big)^2}$ $\Rightarrow\ \text{x}\rightarrow\frac{\pi}{2},$ then $\frac{\pi}{2}-\text{x}\rightarrow0,$ let $\frac{\pi}{2}-\text{x}=\text{y}$ $\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}\rightarrow0}=\lim\limits_{\text{y}\rightarrow0}\frac{\sqrt{2-\sin\big(\frac{\pi}{2}-\text{y}\big)}-1}{\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2-\cos\text{y}}-1}{\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\big(\sqrt{2-\cos\text{y}}-1\big)}{\text{y}^2}\times\frac{\big(\sqrt{2-\cos\text{y}}+1\big)}{\big(\sqrt{2-\cos\text{y}}+1\big)}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\big({2-\cos\text{y}}-1\big)}{\text{y}^2\big(\sqrt{2-\cos\text{y}}+1\big)}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{{1-\cos\text{y}}}{\big(\sqrt{2-\cos\text{y}}+1\big)\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\frac{\text{y}}{2}}{\text{y}^2\big(\sqrt{2-\cos\text{y}}+1\big)}$ $=2\times\Bigg(\lim\limits_{\text{y}\rightarrow0}\frac{\sin\frac{\text{y}}{2}}{\frac{\text{y}}{2}}\Bigg)\times\frac14\frac{1}{\lim\limits_{\text{y}\rightarrow0}\sqrt{2-\cos\text{y}}+1}$ $=2\times1\times\frac14\times\frac{1}{1+1}=\frac14$
View full question & answer→Question 614 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{\text{x}^2-1}+\sqrt{\text{x}-1}}{\sqrt{\text{x}^2-1}},\text{x}>1$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{\text{x}^2-1}+\sqrt{\text{x}-1}}{\sqrt{\text{x}^2-1}}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{\text{x}^2-1}+\sqrt{\text{x}-1}}{\sqrt{\text{x}^2-1}}\times\frac{\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}{\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}\times\frac{\sqrt{\text{x}^2-1}}{\sqrt{\text{x}^-1}}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{\big[\big(\text{x}^2-1\big)-(\text{x}-1)\big]\times\sqrt{\text{x}^2-1}}{\big(\text{x}^2-1\big)\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{\big(\text{x}^2-\text{x}\big)\sqrt{\text{x}^2-1}}{\big(\text{x}^2-1\big)\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}(\text{x}-1)\sqrt{\text{x}^2-1}}{(\text{x}-1)(\text{x}+1)\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}\big(\sqrt{\text{x}-1}\big)\big(\sqrt{\text{x}+1}\big)}{(\text{x}+1)\big(\sqrt{\text{x}-1}\big)\big(\sqrt{\text{x}+1}-1\big)}$ $=\frac{\sqrt{2}}{2\big(\sqrt{2}-1\big)}$ $=\frac{\sqrt{2}}{2\times\big(\sqrt{2}-1\big)}\times\frac{\sqrt{2}+1}{\sqrt{2}+1}$ $=\frac{\sqrt{2}+1}{\sqrt{2}}$
View full question & answer→Question 624 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\tan2\text{x}-\sin2\text{x}}{\text{x}^3}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\tan2\text{x}-\sin2\text{x}}{\text{x}^3}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\sin2\text{x}}{\cos2\text{x}}-\sin2\text{x}}{\text{x}^3}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}\big(\frac{1}{\cos2\text{x}}-1\big)}{\text{x}^3}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}(1-\cos2\text{x})}{\text{x}^2\cos2\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}\big(2\sin^2\text{x}\big)}{\text{x}^3\cos2\text{x}}$ $=\frac{\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}}{\text{x}}\big)\Big(\lim\limits_{\text{x}\rightarrow0}\frac{2\sin^2\text{x}}{\text{x}^2}\Big)}{\big(\lim\limits_{\text{x}\rightarrow0}\cos2\text{x}\big)}$ $=\frac{\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}}{\text{x}}\times2\big)\Big(\lim\limits_{\text{x}\rightarrow0}2\big(\frac{\sin\text{x}}{\text{x}}\big)^2\Big)}{\lim\limits_{\text{x}\rightarrow0}\cos2\text{x}}$ $=\frac{(2\times1)(2\times1)}{1}$ $=4$
View full question & answer→Question 634 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow-2}\frac{\text{x}^3+\text{x}^2+4\text{x}+12}{\text{x}^3+3\text{x}+2}$
Answer$\lim\limits_{\text{x}\rightarrow-2}\frac{\text{x}^3+\text{x}^2+4\text{x}+12}{\text{x}^3+3\text{x}+2}$ $=\lim\limits_{\text{x}\rightarrow-2}\frac{(\text{x}+2)\big(\text{x}^2+\text{x}+6\big)}{(\text{x}+2)\big(\text{x}^2-2\text{x}+1\big)}$ $=\lim\limits_{\text{x}\rightarrow-2}\frac{\text{x}^2+\text{x}+6}{\text{x}^2-2\text{x}+1}$ $=\frac{(-2)^2-(-2)+6}{(-2)^2-2(-2)+1}$ $=\frac{4+2+6}{4+4+1}$ $=\frac{12}{9}$ $=\frac{4}{3}$
View full question & answer→Question 644 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{-1}}\frac{\text{x}^{3}+1}{\text{x}+1}$
Answer$\lim\limits_{\text{x}\rightarrow{-1}}\frac{\text{x}^{3}+1}{\text{x}+1}$$=\lim\limits_{\text{x}\rightarrow{-1}}\frac{\text{x}^3-(-1)^3}{\text{x}-(-1)}$ [Dividing numerator and denominator by x - 1]
Applying formula $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{n}}-\text{a}^\text{n}}{\text{x}-\text{a}}=\text{na}^{\text{n}-1}$
Here, n = 3, a = -1
$\Rightarrow\lim\limits_{\text{x}\rightarrow-1}\frac{\text{x}^3-(-1)^3}{\text{x}-(-1)}=\text{na}^{\text{n}-1}$
$=3(-1)^{3-1}$
$=3(-1)^2$
$=3$
View full question & answer→Question 654 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{a}+\text{x}}-\sqrt{\text{a}}}{\text{x}\sqrt{\text{a}^2+\text{ax}}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{a}+\text{x}}-\sqrt{\text{a}}}{\text{x}\sqrt{\text{a}^2+\text{ax}}}$$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{\text{a}+\text{x}}-\sqrt{\text{a}}\big)}{\text{x}\sqrt{\text{a}^2+\text{ax}}}\times\frac{\big(\sqrt{\text{a}+\text{x}}+\sqrt{\text{a}}\big)}{\big(\sqrt{\text{a}+\text{x}}+\sqrt{\text{a}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(\text{a}+\text{x})-\text{a}}{\text{x}\sqrt{\text{a}^2+\text{ax}}\big(\sqrt{\text{a}+\text{x}}+\sqrt{\text{a}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\text{x}\sqrt{\text{a}^2+\text{ax}}\big(\sqrt{\text{a}+\text{x}}+\sqrt{\text{a}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{1}{\big(\sqrt{\text{a}^2+\text{ax}}\big)\big(\sqrt{\text{a}+\text{x}}+\sqrt{\text{a}}\big)}$
$=\frac{1}{\text{a}\big(\sqrt{2\text{a}}\big)}$
$=\frac{1}{2\text{a}\sqrt{\text{a}}}$
View full question & answer→Question 664 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}+7\text{x}}{4\text{x}+\sin2\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}+7\text{x}}{4\text{x}+\sin2\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\frac{\sin3\text{x}}{\text{x}}+\frac{7\text{x}}{\text{x}}\big)}{\big(\frac{4\text{x}}{\text{x}}+\frac{\sin2\text{x}}{\text{x}}\big)}$ $=\frac{\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{3\text{x}}\times3\Big)+7}{4+\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}}{2\text{x}}\Big)\times2}$ $=\frac{3+7}{4+2}$ $=\frac{10}{6}$ $=\frac53$
View full question & answer→Question 674 Marks
Evaluate the following limit: $\lim\limits_{\text{n}\rightarrow\infty}\frac{\sin\big(\frac{\text{a}}{2^{\text{n}}}\big)}{\sin\big(\frac{\text{b}}{2^{\text{n}}}\big)}$
Answer$\lim\limits_{\text{n}\rightarrow\infty}\frac{\sin\big(\frac{\text{a}}{2^{\text{n}}}\big)}{\sin\big(\frac{\text{b}}{2^{\text{n}}}\big)}$ $\text{n}\rightarrow\infty,$ then $\frac{1}{\text{n}}=\text{h} \rightarrow0$ $=\frac{\lim\limits_{\text{h}\rightarrow\infty}\sin\Bigg(\frac{\text{a}}{2^{\frac{1}{\text{h}}}}\Bigg)}{\lim\limits_{\text{h}\rightarrow\infty}\sin\Bigg(\frac{\text{b}}{2^{\frac{1}{\text{h}}}}\Bigg)}$ $=\frac{\begin{pmatrix}\lim\limits_{\text{h}\rightarrow\infty}\frac{\sin\frac{\text{a}}{2^{\frac{1}{\text{h}}}}}{\frac{\text{a}}{2^{\frac{1}{\text{h}}}}}\times\frac{\text{a}}{2^{\frac{1}{\text{h}}}}\end{pmatrix}}{\begin{pmatrix}\lim\limits_{\text{h}\rightarrow\infty}\frac{\sin\frac{\text{b}}{2^{\frac{1}{\text{h}}}}}{\frac{\text{b}}{2^{\frac{1}{\text{h}}}}}\times\frac{\text{b}}{2^{\frac{1}{\text{h}}}}\end{pmatrix}}$ $=\frac{1\times\frac{\text{a}}{2^{\frac{1}{\text{h}}}}}{1\times\frac{\text{b}}{2^{\frac{1}{\text{h}}}}}=\frac{\text{a}}{\text{b}}$
View full question & answer→Question 684 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\cos2\text{x}-1}{\cos\text{x}-1}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\cos2\text{x}-1}{\cos\text{x}-1}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{1-\cos2\text{x}}{1-\cos\text{x}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin^2\text{x}}{2\sin^2\frac{\text{x}}{2}}$ $=\frac{\lim\limits_{\text{x} \rightarrow0}(\sin\text{x})^2}{\lim\limits_{\text{x} \rightarrow0}\big(\sin\frac{\text{x}}{2}\big)^2}$ $=\frac{\lim\limits_{\text{x} \rightarrow0}\big(2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}\big)^2}{\lim\limits_{\text{x} \rightarrow0}\big(\sin\frac{\text{x}}{2}\big)^2}$ $=4\lim\limits_{\text{x} \rightarrow0}\cos^2\frac{\text{x}}{2}$ $=4\times1$ $=4$
View full question & answer→Question 694 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\text{x}^2}{\sin\pi\text{ x}}$
Answer$\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\text{x}^2}{\sin\pi\text{ x}}$ $\Rightarrow\text{x}\rightarrow1,\text{x}-1\rightarrow0,$ let $\text{x}-1=\text{y}\Rightarrow\text{y}\rightarrow0$ $=\lim\limits_{\text{x}\rightarrow{0}}\frac{1-\text{x}^2}{\sin\pi\text{ x}}=\lim\limits_{\text{x}-1\rightarrow{0}}\frac{(1-\text{x})(1+\text{x})}{\sin\pi\text{ x}}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{(-\text{y})(1+\text{y}+1)}{\sin\pi(\text{y}+1)}$ $=-\lim\limits_{\text{y}\rightarrow{0}}\frac{\text{y}(\text{y}+2)}{\sin\pi(\text{y}+1)}$ $=-\lim\limits_{\text{y}\rightarrow{0}}\frac{\text{y}(\text{y}+2)}{-\sin\pi(\text{y}+1)}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\text{y}(\text{y}+2)}{\frac{\sin\pi\text{y}}{2}}$ $=\frac{\lim\limits_{\text{y}\rightarrow{0}}\text{y}(\text{y}+2)}{\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\pi\text{y}}{\pi\text{y}}\times\pi\text{y}}$ $=\frac{2}{\pi}$
View full question & answer→Question 704 Marks
Evaluate the following limit: Evaluate: $\lim\limits_{\text{n}\rightarrow\infty}\frac{1.2+2.3+3.4+\ \cdots+\text{n}(\text{n}+1)}{\text{n}^3}$
Answer$\lim\limits_{\text{n}\rightarrow\infty}\frac{1.2+2.3+3.4+\ \cdots+\text{n}(\text{n}+1)}{\text{n}^3}$ $=\ \lim\limits_{\text{n}\rightarrow\infty}\frac{\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\frac{\text{n}(\text{n}+1)}{2}}{\text{n}^3}$ $=\lim\limits_{\text{n}\rightarrow\infty}\frac{\text{n}(\text{n}+1)\Big[\frac{(2\text{n}+1)+3}{6}\Big]}{\text{n}^3}$ $=\lim\limits_{\text{n}\rightarrow\infty}\frac{\frac{\text{n}(\text{n}+1)(2\text{n}+4)}{6}}{\text{n}^3}$ $=\lim\limits_{\text{n}\rightarrow\infty}\frac{\Big(1+\frac{1}{\text{n}}\Big)\Big(2+\frac{4}{\text{n}}\Big)}{6}$ $=\frac{1\times2}{6}$ $=\frac{1}{3}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\sqrt{2}-\cos\text{x}-\sin\text{x}}{\big(\frac{\pi}{4}-\text{x}\big)^2}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\sqrt{2}-\cos\text{x}-\sin\text{x}}{\big(\frac{\pi}{4}-\text{x}\big)^2}$As $\text{x}\rightarrow\frac{\pi}{4},\frac\pi4-\text{x}\rightarrow0,$ let $\frac{\pi}{4}-\text{x}=\text{y}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\cos\big(\frac\pi4-\text{y}\big)-\sin\big(\frac\pi4-\text{y}\big)}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\big[\cos\frac\pi4\cos\text{y}+\sin\frac{\pi}{4}\sin\text{y}+\sin\frac\pi4\cos\text{y}-\cos\frac\pi4\sin\text{y}\big]}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\Big(\frac{\cos\text{y}}{\sqrt{2}}+\frac{\sin\text{y}}{\sqrt{2}}+\frac{\cos\text{y}}{\sqrt{2}}-\frac{\sin\text{y}}{\sqrt{2}}\Big)}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\frac{2\cos\text{y}}{\sqrt{2}}}{\text{y}^2}=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\sqrt{2}\cos\text{y}}{\text{y}^2}$
$=\sqrt{2}\lim\limits_{\text{y}\rightarrow{0}}\frac{(1-\cos\text{y})}{\text{y}^2}$
$=\sqrt{2}\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\frac{\text{y}}{2}}{\frac{\text{y}^2}{4}}\times\frac14$
$=\sqrt{2}\times2\times\frac14\bigg(\lim\limits_{\text{y}\rightarrow0}\frac{\sin\frac{\text{y}}{2}}{\frac{\text{y}}{2}}\bigg)^2$
$=\sqrt{2}\times2\times\frac14\times1$
$=\frac{1}{\sqrt{2}}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{3\sin^2\text{x}-2\sin\text{x}^2}{3\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{3\sin^2\text{x}-2\sin\text{x}^2}{3\text{x}^2}$$=\lim\limits_{\text{x} \rightarrow0}\frac{3\sin^2\text{x}-2\sin\text{x}^2}{3\text{x}^2}-\lim\limits_{\text{x} \rightarrow0}\frac{2\sin\text{x}^2}{3\text{x}^2}$
$=\lim\limits_{\text{x} \rightarrow0}\big(\frac{\sin\text{x}}{\text{x}}\big)^2-\frac{2}{3}\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}^2}{\text{x}^2}$
$=1-\frac{2}{3}\times1$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=1-\frac23$
$=\frac{3-2}{3}$
$=\frac{1}{3}$
View full question & answer→Question 734 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow2}\frac{\sqrt{1+4\text{x}}+\sqrt{5+2\text{x}}}{\text{x}-2}$
Answer$\lim\limits_{\text{x}\rightarrow2}\frac{\sqrt{1+4\text{x}}+\sqrt{5+2\text{x}}}{\text{x}-2}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{\big(\sqrt{1+4\text{x}}-\sqrt{5+2\text{x}}\big)}{(\text{x}-2)}\times\frac{\big(\sqrt{1+4\text{x}}+\sqrt{5+2\text{x}}\big)}{\big(\sqrt{1+4\text{x}}+\sqrt{5+2\text{x}}\big)}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{(1+4\text{x})-(5+2\text{x})}{(\text{x}-2)\big(\sqrt{1+4\text{x}}+\sqrt{5+2\text{x}}\big)}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{-4+2\text{x}}{(\text{x}-2)\big(\sqrt{1+4\text{x}}+\sqrt{5+2\text{x}}\big)}$ $=\frac{2}{\sqrt{1+8}+\sqrt{5+4}}=\frac{2}{\sqrt{9}+\sqrt{9}}$ $=\frac{2}{6}=\frac13$
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Evaluate the following limit: $\lim\limits_{\text{n}\rightarrow\infty}\frac{\text{x}^4+7\text{x}^3+46\text{x}+\text{a}}{\text{x}^4+6},$ where a is a non-zero real number.
Answer$\lim\limits_{\text{n}\rightarrow\infty}\frac{\text{x}^4+7\text{x}^3+46\text{x}+\text{a}}{\text{x}^4+6}$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$ $=\lim\limits_{\text{n}\rightarrow\infty}\frac{1+\frac{7}{\text{x}}+\frac{46}{\text{x}^3}+\frac{\text{a}}{\text{x}^4}}{1+\frac{6}{\text{x}^4}}$ $=1$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}(1-\text{x})\tan\Big(\frac{\pi\text{x}}{2}\Big)$
Answer$\lim\limits_{\text{x}\rightarrow1}(1-\text{x})\tan\Big(\frac{\pi\text{x}}{2}\Big)$ When x → 1, x - 1 → 0, let x - 1 = y, then y → 0 $=\lim\limits_{{(\text{x}-1)\rightarrow0}}-(\text{x}-1)\tan\frac{\pi\text{x}}{2}$ $=-\lim\limits_{\text{y}\rightarrow0}\text{y}\tan\frac{\text{y}}{2}(\text{y}+1)$ $=-\lim\limits_{\text{y}\rightarrow0}\text{y}\times\tan\Big(\frac{\pi}{2}+\frac{\pi}{2}\text{y}\Big)$ $=\lim\limits_{\text{y}\rightarrow0}\text{y}\times\cot\frac{\pi}{2}\text{y}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{\text{y}}{\tan\frac{\pi\text{y}}{2}}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{\frac{\pi\text{y}}{2}\times\frac{2}{\pi}}{\tan\frac{\pi\text{y}}{2}}$ $=\frac2\pi$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{4}}\frac{\text{x}^{3}-64}{\text{x}^2-16}$
Answer$\lim\limits_{\text{x}\rightarrow{4}}\frac{\text{x}^{3}-64}{\text{x}^2-16}$$=\lim\limits_{\text{x}\rightarrow{4}}\frac{\text{x}^{3}-4^3}{\text{x}^2-4^2}$
$=\lim\limits_{\text{x}\rightarrow{4}}\frac{\frac{\text{x}^{3}-4^3}{\text{x}-4}}{\frac{\text{x}^2-4^2}{\text{x}-4}}$ [Dividing numerator and denominator by x - 4]
$=\frac{\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}^3-4^3}{\text{x}-4}}{\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}^2-4}{\text{x}-4}}$
Applying formula $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{n}}-\text{a}^\text{n}}{\text{x}-\text{a}}=\text{na}^{\text{n}-1}$ in numerator and $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{m}}-\text{a}^{\text{m}}}{\text{x}-\text{a}}=\text{ma}^{\text{m}-1}$ in denominator
$\Rightarrow\text{n}=3,\text{m}=2$
$\Rightarrow\frac{\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}^3-4^3}{\text{x}-4}}{\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}^2-4}{\text{x}-4}}=\frac{3(4)^{3-1}}{2(4)^{2-1}}=\frac{3(4)^2}{2(4)}$
$=6$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sin(3+\text{x})-\sin(3-\text{x})}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin(3+\text{x})-\sin(3-\text{x})}{\text{x}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{2\cos\big(\frac{3+\text{x}+3-\text{x}}{2}\big)\sin\big(\frac{3+\text{x}-3+\text{x}}{2}\big)}{\text{x}}$ $=2\lim\limits_{\text{x} \rightarrow0}\frac{\cos3.\sin\text{x}}{\text{x}}$ $=2\cos3\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}}{\text{x}}$ $=2\cos3\times1$ $=2\cos3$
View full question & answer→Question 784 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow-\infty}\big(\sqrt{\text{x}^2-8\text{x}}+\text{x}\big)$
Answer$\lim\limits_{\text{x}\rightarrow-\infty}\big(\sqrt{\text{x}^2-8\text{x}}+\text{x}\big)$ $=\lim\limits_{\text{y}\rightarrow\infty}\Big(\sqrt{\text{y}^2+8\text{y}}-\text{y}\Big),$ where y = -x on rationalising $=\lim\limits_{\text{y}\rightarrow\infty}\frac{\big(\sqrt{\text{y}^2+8\text{y}}-\text{y}\big)\big(\sqrt{\text{y}^2+8\text{y}}+\text{y}\big)}{\big(\sqrt{\text{y}^2+8\text{y}}+\text{y}\big)}$ $=\lim\limits_{\text{y}\rightarrow\infty}\frac{\text{y}^2+8\text{y}+\text{y}^2}{\sqrt{\text{y}^2+8\text{y}}+\text{y}}$ $=\lim\limits_{\text{y}\rightarrow\infty}\frac{8\text{y}}{\text{y}\sqrt{1+\frac{8}{\text{y}}}+\text{y}}$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$ $=\lim\limits_{\text{y}\rightarrow\infty}\frac{8}{\sqrt{1+\frac{8}{\text{y}}}+1}$ $=\frac{8}{1+1}=\frac{8}{2}$ $=4$
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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$\lim\limits_{\text{h}\rightarrow0}\frac{\text{(a}+\text{h})^2\sin(\text{a}+\text{h})-\text{a}^2\sin\text{a}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\big(\text{a}^2+\text{h}^2+2\text{ah}\big)\sin(\text{a}+\text{h})-\text{a}^2\sin\text{a}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}^2\sin(\text{a}+\text{h})+\text{h}^2\sin(\text{a}+\text{h})+2\text{ah}\sin(\text{a}+\text{h})-\text{a}^2\sin\text{a}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\Big[\frac{\text{a}^2(\sin(\text{a}+\text{h})-\sin\text{a})}{\text{h}}+\frac{\text{h}^2\sin(\text{a}+\text{h})}{\text{h}(\text{a}+\text{h})}\times(\text{a}+\text{h})+\frac{2\text{ah}}{\text{h}}(\sin(\text{a}+\text{h}))\Big]$ $=\Bigg[\text{a}^2\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\big(\frac{\text{a}+\text{h}+\text{a}}{2}\big)\sin\big(\frac{\text{a}+\text{h}-\text{a}}{2}\big)}{\text{h}}\Bigg]+[0]+2\text{a}\lim\limits_{\text{h}\rightarrow0}\sin(\text{a}+\text{h})$ $=\Big(2\text{a}^2\cos\text{a}\times\frac12\Big)+(2\text{a}\sin\text{a})$ $=\text{a}^2\cos\text{a}+2\text{a}\sin\text{a}$
View full question & answer→Question 804 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow4}\frac{2-\sqrt{\text{x}}}{4-\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow4}\frac{2-\sqrt{\text{x}}}{4-\text{x}}$$\lim\limits_{\text{x}\rightarrow4}\frac{\big(2-\sqrt{\text{x}}\big)\big(2+\sqrt{\text{x}}\big)}{(4-\text{x})\big(2+\sqrt{\text{x}}\big)}$
$\lim\limits_{\text{x}\rightarrow4}\frac{(2)^2-\big(\sqrt{\text{x}}\big)^2}{(4-\text{x})\big(2+\sqrt{\text{x}}\big)}$
$\lim\limits_{\text{x}\rightarrow4}\frac{(4-\text{x})}{(4-\text{x})\big(2+\sqrt{\text{x}}\big)}$
$=\frac{1}{2+\sqrt{4}}$
$=\frac{1}{2+2}$
$=\frac14$
View full question & answer→Question 814 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{2}-\sqrt{1+\cos\text{x}}}{\sin^2\text{x}}$
AnswerGiven that: $\lim\limits_{\text{x} \rightarrow0}\frac{\sqrt{2}-\sqrt{1+\cos\text{x}}}{\sin^{2}\text{x}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{\sqrt{2}-\sqrt{1+\cos\text{x}}}{\sin^{2}\text{x}}\times\frac{\sqrt{2}+\sqrt{1+\cos\text{x}}}{\sqrt{2}+\sqrt{1+\cos\text{x}}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{2-(1+\cos\text{x})}{\sin^{2}\text{x}\big[\sqrt{2}+\sqrt{1+\cos\text{x}}\big]}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{1+\cos\text{x}}{\sin^{2}\text{x}\Big[\sqrt{2}+\sqrt{1+\cos\text{x}}\Big]}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin^{2}\frac{\text{x}}{2}}{(2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2})^{2}}\times\frac{1}{\sqrt{2}+\sqrt{1+\cos\text{x}}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin^{2}\frac{\text{x}}{2}}{4\sin^{2}\frac{\text{x}}{2}\cos^{2}\frac{\text{x}}{2}}\times\frac{1}{\sqrt{2}+\sqrt{1+\cos\text{x}}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{2}{4\cos^{2}\frac{\text{x}}{2}}\times\frac{1}{\sqrt{2}+\sqrt{1+\cos\text{x}}}$ Taking limit, we get: $=\frac{2}{4\cos^{2}0}\times\frac{1}{(\sqrt{2}+\sqrt{2})}$ $=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2\sqrt{2}}=\frac{1}{4\sqrt{2}}$ Hence, the required answer is $\frac{1}{4\sqrt{2}}.$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{1}}\frac{1+\cos\pi\text{x}}{(1-\text{x})^2}$
Answer$\lim\limits_{\text{x}\rightarrow{1}}\frac{1+\cos\pi\text{x}}{(1-\text{x})^2}$ $\Rightarrow\text{x}\rightarrow1,\text{x}-1\rightarrow0,$ let $\text{x}-1=\text{y}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{1+\cos\pi(\text{y}+1)}{(-\text{y})^2}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{1+\cos(\pi+\pi\text{y})}{\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{1-\cos(\pi\text{y})}{\text{y}^2}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{2-\sin^2\frac{\pi\text{y}}{2}}{\text{y}^2}$ $=2\Bigg(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\frac{\pi\text{y}}{2}}{\frac{\pi\text{y}}{2}}\Bigg)^2\times\frac{\pi^2}{4}$ $=2\times1\times\frac{\pi^2}{4}$ $=\frac{\pi^2}{2}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\sin\sqrt{\text{x}}-\sin\sqrt{\text{a}}}{\text{x}-\text{a}}$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\sin\sqrt{\text{x}}-\sin\sqrt{\text{a}}}{\text{x}-\text{a}}$$=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\sin\sqrt{\text{x}}-\sin\sqrt{\text{a}}}{\big(\sqrt{\text{x}}-\sqrt{\text{a}}\big)\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)}$
$=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{2\sin\Big(\frac{\sqrt{\text{x}}-\sqrt{\text{a}}}{2}\Big)\cos\Big(\frac{\sqrt{\text{x}}+\sqrt{\text{a}}}{2}\Big)}{\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)\Big({\sqrt{\text{x}}-\sqrt{\text{a}}}\Big)}$
$=2\begin{pmatrix}\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\sin\frac{\sqrt{\text{x}}-\sqrt{\text{a}}}{2}}{\Big(\frac{{\sqrt{\text{x}}+\sqrt{\text{a}}}}{2}\Big)}\end{pmatrix}\times\frac12\frac{\lim\limits_{\text{x}\rightarrow{\text{a}}}\cos\Big(\frac{\sqrt{\text{x}}+\sqrt{\text{a}}}{2}\Big)}{\lim\limits_{\text{x}\rightarrow{\text{a}}}\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)}$
$=2\times1\times\frac{1}{2}\times\cos\sqrt{\text{a}}\times\frac{1}{2\sqrt{\text{a}}}$
$=\frac{\cos\sqrt{\text{a}}}{2\sqrt{\text{a}}}$
View full question & answer→Question 844 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}-\sin2\text{x}}{\text{x}^3}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}-\sin2\text{x}}{\text{x}^3}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}-2\sin\text{x}\cos\text{x}}{\text{x}^3}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}(1-\cos\text{x})}{\text{x}^3}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}(1-\cos\text{x})}{\text{x}^3}\times\frac{1+\cos\text{x}}{1+\cos\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\big(1-\cos^2\text{x}\big)}{\text{x}^3(1+\cos\text{x})}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\big(\sin^2\text{x}\big)}{\text{x}^3(1+\cos\text{x})}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin^3\text{x}}{\text{x}^3(1+\cos\text{x})}$ $=2\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\sin\text{x}}{\text{x}}\Big)^3\times\lim\limits_{\text{x}\rightarrow0}\frac{1}{(1+\cos\text{x})}$ $=2\times1\times\frac{1}{(1+1)}$ $=1$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\sqrt{\cos\text{x}}-\sqrt{\sin\text{x}}}{\text{x}-\frac{\pi}{4}}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\sqrt{\cos\text{x}}-\sqrt{\sin\text{x}}}{\text{x}-\frac{\pi}{4}}$$=\lim\limits_{{\text{x}-{\frac\pi4\rightarrow0}}}\frac{\big(\sqrt{\cos\text{x}}-\sqrt{\sin\text{x}}\big)}{\big(\text{x}-\frac{\pi}{4}\big)}\times\frac{\big(\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}\big)}{\big(\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}\big)}$
$=\lim\limits_{{\text{x}-{\frac\pi4\rightarrow0}}}\frac{\big(\sqrt{\cos\text{x}}-\sqrt{\sin\text{x}}\big)}{\big(\text{x}-\frac{\pi}{4}\big)\big(\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}\big)}$
As $\text{x}\rightarrow\frac{\pi}{4}\Rightarrow\text{x}-\frac\pi4\rightarrow0\Rightarrow$let $\text{x}-\frac\pi4=\text{y}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{\Big(\cos\big(\frac\pi4+\text{y}\big)-\sin\big(\frac\pi4+\text{y}\big)\Big)}{\text{y}\Big(\sqrt{\cos\big(\frac\pi4+\text{y}\big)}+\Big(\sqrt{\sin\big(\frac\pi4+\text{y}\big)}\Big)}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{\big(\cos\frac\pi4\cos\text{y}-\sin\frac{\pi}{4}\sin\text{y}\big)-\big(\sin\frac\pi4\cos\text{y}-\cos\frac{\pi}{4}\sin\text{y}\big)}{\text{y}\Big(\sqrt{\cos\big(\frac\pi4+\text{y}\big)}+\Big(\sqrt{\sin\big(\frac\pi4+\text{y}\big)}\Big)}$
$=\lim\limits_{\text{y}\rightarrow0}\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(\sqrt{\cos\big(\frac\pi4+\text{y}\big)}+\Big(\sqrt{\sin\big(\frac\pi4+\text{y}\big)}\Big)}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{\Big(-2\frac{\sin\text{y}}{\sqrt{2}}\Big)}{\text{y}\Big(\sqrt{\cos\big(\frac\pi4+\text{y}\big)}+\Big(\sqrt{\sin\big(\frac\pi4+\text{y}\big)}\Big)}$
$=-\sqrt{2}\Big(\lim\limits_{\text{y}\rightarrow0}\frac{\sin\text{y}}{\text{y}}\Big)\times\frac{1}{\lim\limits_{\text{y}\rightarrow0}\sqrt{\cos\big(\text{y}+\frac\pi4}\big)+\lim\limits_{\text{y}\rightarrow0}\sqrt{\sin\big(\text{y}+\frac{\pi}{4}\big)}}$
$=-\sqrt{2}\times1\times\frac{1}{\sqrt{\cos\frac{\pi}{4}}+\sqrt{\sin\frac\pi4}}$
$=-\sqrt{2}\times\frac{1}{\big(\frac{1}{\sqrt{2}}\big)^{\frac{1}{2}}+\big(\frac{1}{\sqrt{2}}\big)^{\frac{1}{2}}}$ $\Big[\because\cos\frac\pi4=\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}\Big]$
$=\frac{-\sqrt{2}}{\big(\frac{1}{\sqrt{2}}\big)^{\frac{1}{2}}+(1+1)^{\frac{1}{2}}}$
$=\frac{-\sqrt{2}}{\sqrt{2}\big(\frac{1}{\sqrt{2}}\big)^{\frac{1}{2}}}$
$=-\frac{1}{2^{\frac14}}$
View full question & answer→Question 864 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{3+\text{x}}-\sqrt{5-\text{x}}}{\text{x}^2-1}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{3+\text{x}}-\sqrt{5-\text{x}}}{\text{x}^2-1}$$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{3+\text{x}}+\sqrt{5-\text{x}}\big)}{(\text{x}-1)(\text{x}+1)}\times\frac{\big(\sqrt{3+\text{x}}+\sqrt{5-\text{x}}\big)}{\big(\sqrt{3+\text{x}}+\sqrt{5-\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{((3+\text{x})-(5-\text{x}))}{(\text{x}-1)(\text{x}+1)\big(\sqrt{3+\text{x}}+\sqrt{5-\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{-2+2\text{x}}{(\text{x}-1)(\text{x}+1)\big(\sqrt{3+\text{x}}+\sqrt{5-\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{-2(\text{x}-1)}{(\text{x}-1)(\text{x}+1)\big(\sqrt{3+\text{x}}+\sqrt{5-\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{-2}{(\text{x}+1)\big(\sqrt{3+\text{x}}+\sqrt{5-\text{x}}\big)}$
$=\frac{2}{(1+1)\big(\sqrt{3+1}+\sqrt{5-1}\big)}$
$=\frac{2}{(2)(2+2)}$
$=\frac14$
View full question & answer→Question 874 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^2-\text{x}-2}{\text{x}^2-2\text{x}+\sin(\text{x}-2)}$
Answer$\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^2-\text{x}-2}{\text{x}^2-2\text{x}+\sin(\text{x}-2)}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}-2)(\text{x}+1)}{\text{x}^2-2\text{x}+\sin(\text{x}-2)}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{1}{\frac{\text{x}}{\text{x}+1}+\frac{\sin(\text{x}-2)}{(\text{x}-2)(\text{x}+1)}}$ $=\lim\limits_{\text{x}\rightarrow2}(\text{x}+1)\Bigg(\frac{1}{\text{x}+\frac{\sin(\text{x}-2)}{\text{x}-2}}\Bigg)$ $=\lim\limits_{\text{x}\rightarrow2}(\text{x}+1)\frac{1}{\lim\limits_{\text{x}\rightarrow2}(\text{x})+\lim\limits_{\text{x}\rightarrow2}\frac{\sin(\text{x}-2)}{\text{x}-2}}$ $=(2+1)\times\frac{1}{(2)\lim\limits_{\text{x}\rightarrow2\rightarrow0}\frac{\sin(\text{x}-2)}{(\text{x}-2)}}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$ $=3\times\frac{1}{2+1}$ $=1$
View full question & answer→Question 884 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{ax}+\text{bx}}{\text{ax}+\sin\text{bx}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{ax}+\text{bx}}{\text{ax}+\sin\text{bx}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\sin\text{ax}}{\text{x}}+\text{b}}{\text{a}+\frac{\sin\text{bx}}{\text{bx}}}$ $=\frac{\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{ax}}{\text{ax}}\times\text{a}+\text{b}}{\text{a}+\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{bx}}{\text{bx}}\times\text{b}}$ $=\frac{\text{a}+\text{b}}{\text{a}+\text{b}}$ $=1$
View full question & answer→Question 894 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{{\text{x}-1}}{\sqrt{\text{x}^2+3}-2}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{{\text{x}-1}}{\sqrt{\text{x}^2+3}-2}$$=\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}-1)\times\big(\sqrt{\text{x}^2+3}+2\big)}{\big(\sqrt{\text{x}^2+3}-2\big)\big(\sqrt{\text{x}^2+3}+2\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}-1)\big(\sqrt{\text{x}^2+3}+2\big)}{\big({\text{x}^2+3}-4\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}-1)\big(\sqrt{\text{x}^2+3}+2\big)}{\big({\text{x}^2-1}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{\text{x}^2+3}+2}{{\text{x}}+1}$
Putting the value x = 1
$\Rightarrow\frac{\sqrt{1+3}+2}{1+1}$
$=\frac{2+2}{2}$
$=\frac{4}{2}=2$
View full question & answer→Question 904 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow2}\frac{\sqrt{3+\text{x}}-1}{2-\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow2}\frac{\sqrt{3+\text{x}}-1}{2-\text{x}}$$=\lim\limits_{\text{x}\rightarrow2}\frac{\big(\sqrt{3-\text{x}}-1\big)}{2-\text{x}}\times\frac{\big(\sqrt{3-\text{x}}+1\big)}{\big(\sqrt{3-\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{(3-\text{x})-1}{(2-\text{x})\big(\sqrt{3-\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{(2-\text{x})}{(2-\text{x})\big(\sqrt{3-\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{1}{\big(\sqrt{3-\text{x}}+1\big)}$
$=\frac{1}{\sqrt{3-2}+1}=\frac{1}{1+1}$
$=\frac{1}{2}$
View full question & answer→Question 914 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{{\text{x}}-\text{a}}{\sqrt{\text{x}}-\sqrt{\text{a}}}$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{{\text{x}}-\text{a}}{\sqrt{\text{x}}-\sqrt{\text{a}}}$$=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{({\text{x}}-\text{a})\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)}{\big(\sqrt{\text{x}}-\sqrt{\text{a}}\big)\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)}$
$=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{(\text{x}-\text{a})-\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)}{(\text{x}-\text{a})}$
$=\lim\limits_{\text{x}\rightarrow{\text{a}}}\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)$
$=2\sqrt{\text{a}}$
View full question & answer→Question 924 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{(\text{x}+2)^{\frac{3}{2}}-(\text{a}+2)^{\frac{3}{2}}}{\text{x}-\text{a}}$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{(\text{x}+2)^{\frac{3}{2}}-(\text{a}+2)^{\frac{3}{2}}}{\text{x}-\text{a}}$ $=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{(\text{x}+2)^{\frac{3}{2}}-(\text{a}+2)^{\frac{3}{2}}}{(\text{x}+2)-(\text{a}+2)}$ Let x + 2 = y, a + 2 = b $\Rightarrow\lim\limits_{({\text{x}+2)}\rightarrow{(\text{a}+2)}}\frac{(\text{y})^\frac{3}{2}-(\text{b})^\frac{3}{2}}{(\text{y})-(\text{b})},$ $\Big[\text{Using 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}\Big]$ $=\frac{3}{2}(\text{b})^{\frac{3}{2}-1}$ $=\frac{3}{2}(\text{a}+2)^{\frac{3}{2}-1}$ $=\frac{3}{2}(\text{a}+2)^{\frac{1}{2}}$
View full question & answer→Question 934 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\cos\sqrt{\text{x}}-\cos\sqrt{\text{a}}}{\text{x}-\text{a}}$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\cos\sqrt{\text{x}}-\cos\sqrt{\text{a}}}{\text{x}-\text{a}}$ $=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{-2\sin\Big(\frac{\sqrt{\text{x}}+\sqrt{\text{a}}}{2}\Big)\times\sin\Big(\frac{\sqrt{\text{x}}-\sqrt{\text{a}}}{2}\Big)}{\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)\times\Big({\sqrt{\text{x}}-\sqrt{\text{a}}}\Big)}$ $=-2\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\sin\Big(\frac{\sqrt{\text{x}}+\sqrt{\text{a}}}{2}\Big)\times\lim\limits_{\text{x}\rightarrow{\text{a}}}\sin\Big(\frac{\sqrt{\text{x}}-\sqrt{\text{a}}}{2}\Big)}{\lim\limits_{\text{x}\rightarrow{\text{a}}}\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)\times\Big(\frac{\sqrt{\text{x}}-\sqrt{\text{a}}}{2}\Big)}\times\frac12$ $=-2\sin\sqrt{\text{a}}\times1\times\frac{1}{2\sqrt{\text{a}}}\times\frac12$ $=-\frac{1}{2\sqrt{\text{a}}}\sin\sqrt{\text{a}}$
View full question & answer→Question 944 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}-3}{\sqrt{\text{x}-2}-\sqrt{4-\text{x}}}$
Answer$\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}-3}{\sqrt{\text{x}-2}-\sqrt{4-\text{x}}}$$=\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}-3}{\big(\sqrt{\text{x}-2}-\sqrt{4-\text{x}}\big)}\times\frac{\big(\sqrt{\text{x}}-2+\sqrt{4-\text{x}}\big)}{\big(\sqrt{\text{x}}-2+\sqrt{4-\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}-3)\big(\sqrt{\text{x}-2}+\sqrt{4-\text{x}}\big)}{(\text{x}-2)-(4-\text{x})}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}-3)\big(\sqrt{\text{x}-2}+\sqrt{4-\text{x}}\big)}{\text{x}-2-4+\text{x}}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}-3)\big(\sqrt{\text{x}-2}+\sqrt{4-\text{x}}\big)}{2(\text{x}-3)}$
$=\frac{1}{2}\lim\limits_{\text{x}\rightarrow3}\big(\sqrt{\text{x}-2}+\sqrt{4-\text{x}}\big)$
$=\frac{1}{2}\big(\sqrt{3-2}+\sqrt{4-\text{x}}\big)$
$=\frac{1}{2}\big(\sqrt{1}+\sqrt{1}\big)$
$=\frac{1}{2}(1+1)=\frac22$
$=1$
View full question & answer→Question 954 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\infty}}{\sqrt{\text{x}^2+\text{cx}}-\text{x}}{}$
Answer$\lim\limits_{\text{x}\rightarrow{\infty}}{\sqrt{\text{x}^2+\text{cx}}-\text{x}}{}$$=\lim\limits_{\text{x}\rightarrow{\infty}}\Bigg(\big(\sqrt{\text{x}^2+\text{cx}}-\text{x}\big)\frac{\big(\sqrt{\text{x}^2+\text{cx}}+\text{x}\big)}{\sqrt{\text{x}^2+\text{cx}}+\text{x}}\Bigg)$
$=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{\big(\text{x}^2+\text{cx}-\text{x}^2\big)}{\sqrt{\text{x}^2+\text{cx}+\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{\text{cx}}{\sqrt{\text{x}^2+\text{cx}+\text{x}}}$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$$=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{\text{c}}{\sqrt{1+\frac{\text{c}}{\text{x}}+1}}$
$=\frac{\text{c}}{1+1}=\frac{\text{c}}{2}$
View full question & answer→Question 964 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{5\text{x}-4}-\sqrt{\text{x}}}{\text{x}-1}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{5\text{x}-4}-\sqrt{\text{x}}}{\text{x}-1}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{\big(\sqrt{5\text{x}-4}-\sqrt{\text{x}}\big)}{\text{x}-1}\times\frac{\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}{\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}{}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{((5\text{x}-4)-\text{x)}}{(\text{x}-1)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$ $=4\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}-1)}{(\text{x}-1)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$ $=4\lim\limits_{\text{x}\rightarrow1}\frac{1}{\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$ $=4\times\frac{1}{\sqrt{5-4}+\sqrt{1}}$ $=4\times\frac{1}{\sqrt{5-4}+\sqrt{1}}$ $=4\times\frac{1}{\sqrt{1}+\sqrt{1}}$ $=\frac{4}{2}=2$
View full question & answer→Question 974 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{7}}-\text{a}^{\frac{2}{7}}}{\text{x}-\text{a}}$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{7}}-\text{a}^{\frac{2}{7}}}{\text{x}-\text{a}}$Applying formula $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{n}}-\text{a}^{\text{n}}}{\text{x}-\text{a}}=\text{na}^{\text{n}-1},$ here, $\text{n}=\frac27$
$\Rightarrow\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{7}}-\text{a}^{\frac{2}{7}}}{\text{x}-\text{a}}=\frac{2}{7}(\text{a})^{\frac{2}{7}-1}$
$=\frac{2}{7}\text{a}^{\frac{-5}{7}}$
View full question & answer→Question 984 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{5\text{x}-4}-\sqrt{\text{x}}}{\text{x}^3-1}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{5\text{x}-4}-\sqrt{\text{x}}}{\text{x}^3-1}$$=\lim\limits_{\text{x}\rightarrow1}\frac{\big(\sqrt{5\text{x}-4}-\sqrt{\text{x}}\big)}{(\text{x}-1)\big(\text{x}^2+1+1\big)}\times\frac{\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}{\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{(5\text{x}-4-\text{x})}{(\text{x}-1)\big(\text{x}^2+1+\text{x}\big)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{4(\text{x}-1)}{(\text{x}-1)\big(\text{x}^2+\text{x}+1\big)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=\frac{4}{(1+1+1)\big(\sqrt{5-4}+\sqrt{1}\big)}$
$=\frac{4}{(3)(1+1)}$
$=\frac{4}{3\times2}=\frac23$
View full question & answer→Question 994 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos4\text{x}}{\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos4\text{x}}{\text{x}^2}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin^22\text{x}}{\text{x}^2}$ $=2\lim\limits_{\text{x} \rightarrow0}\Big(\frac{\sin2\text{x}}{\text{x}}\Big)^2$ $=2\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\sin2\text{x}}{2\text{x}}\Big)^2\times(2)^2$ $=2\times1\times4$ $=8$
View full question & answer→Question 1004 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^{\text{n}}-3^\text{n}}{\text{x}-\text{3}}=108,$ find the value of n.
Answer$\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^{\text{n}}-3^\text{n}}{\text{x}-\text{3}}=108$$\text{L.H.S}=\lim\limits_{\text{x}\rightarrow{3}}\frac{\text{x}^{\text{n}}-3^{\text{n}}}{\text{x}-3}$
Applying the formula $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{n}}-\text{a}^\text{n}}{\text{x}-\text{a}}=\text{na}^{\text{n}-1}$
Here, n = n, a = 3
$\Rightarrow{\lim\limits_{\text{x}\rightarrow{3}}}\frac{\text{x}^{\text{n}}-3^\text{n}}{\text{x}-3}=\text{n}(3)^{\text{n}-1}$
It is given that $\text{n}(3)^{\text{n}-1}=108$
$\Rightarrow\text{n}(3)^{\text{n}-1}=2\times2\times3\times3\times3$
$=(2)^2\times(3)^3$
$=4(3)^{4-1}$
$\Rightarrow\text{n}=4$
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