Question 13 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\sqrt{2}-\sqrt{1+\sin\text{x}}}{\cos^2\text{x}}$
Answer $\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\sqrt{2}-\sqrt{1+\sin\text{x}}}{\cos^2\text{x}}$ $=\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\sqrt{2}-\sqrt{1+\sin\text{x}}}{\cos^2\text{x}}\frac{\sqrt{2}+\sqrt{1+\sin\text{x}}}{\sqrt{2}+\sqrt{1+\sin\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{2-1-\sin\text{x}}{\cos^2\text{x}\big(\sqrt{2}-\sqrt{1+\sin\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{1-\sin\text{x}}{\big(1-\sin^2\text{x}\big)\big(\sqrt{2}-\sqrt{1+\sin\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{1}{\big(1+\sin\text{x}\big)\big(\sqrt{2}+\sqrt{1+\sin\text{x}}\big)}$
$=\frac{1}{(1+1)\big(\sqrt{2}+\sqrt{2}\big)}$
$=\frac{1}{\big(4\sqrt{2}\big)}$
View full question & answer→Question 23 Marks
Let f(x) be a function defined by $\text{f(x)}=\begin{cases}\frac{3\text{x}}{|\text{x}|+2\text{x}}, & \text{x} \neq0\\\ \ \ \ 0, & \text{x} = 0\end{cases}.$
Answer$\lim\limits_{\text{x}\rightarrow0^{-}}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0^{-}}\frac{3\text{x}}{-\text{x}+2\text{x}}=\lim\limits_{\text{x}\rightarrow0^{-}}\frac{3\text{x}}{\text{x}}=3$ $[\because \text{x}\rightarrow^-,|\text{x}|=-\text{x}]$ $\lim\limits_{\text{x}\rightarrow0^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0^+}\frac{3\text{x}}{\text{x}+2\text{x}}=1$ $[\because \text{x}\rightarrow0^+,|\text{x}|=\text{x}]$ Thus, $\lim\limits_{\text{x}\rightarrow0^{-}}\text{f(x)}\neq\lim\limits_{\text{x}\rightarrow0^{+}}\text{f(x)}$ $\therefore\ \lim\limits_{\text{x}\rightarrow0}\text{f(x)}$ does not exist.
View full question & answer→Question 33 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}^2\big(1-\cos\text{x}^2\big)}{\text{x}^6}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}^2\big(1-\cos\text{x}^2\big)}{\text{x}^6}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}^2\times2\sin^2\frac{\text{x}^2}{2}}{\text{x}^6}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}^2}{\text{x}^2}\times\lim\limits_{\text{x}\rightarrow0}\frac{2\sin^2\frac{\text{x}^2}{2}}{\text{x}^4}$
$=\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}^2}{\text{x}}\Big)\times2\times\Bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\frac{\text{x}^2}{2}}{\frac{\text{x}^2}{2}}\Bigg)\times\frac14$
$=(1)^2\times2\times1\times\frac14$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=\frac12$
View full question & answer→Question 43 Marks
Evaluate the following limit:
Show that $\lim\limits_{\text{x}\rightarrow\infty}\big(\sqrt{\text{x}^2+\text{x}+1}-\text{x}\big)\ne\lim\limits_{\text{x}\rightarrow\infty}\big(\sqrt{\text{x}^2+1}-\text{x}\big)$
Answer$\text{R.H.S=}\lim\limits_{\text{x}\rightarrow\infty}\big(\sqrt{\text{x}^2+1}-\text{x}\big)$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\big(\sqrt{\text{x}^2+1}-\text{x}\big)\big(\sqrt{\text{x}^2+1}+\text{x}\big)}{\big(\sqrt{\text{x}^2+1}-\text{x}\big)}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{{\text{x}^2+1}-\text{x}}{\text{x}\sqrt{1+\frac{1}{\text{x}^2}}+\text{x}}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{1}{\text{x}\sqrt{1+\frac{1}{\text{x}^2}}+\text{x}}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{1}{\text{x}}\begin{pmatrix}\frac{1}{\text{x}\sqrt{1+\frac{1}{\text{x}^2}}+\text{x}}\end{pmatrix}$
$=0$
Also,
$\text{L.H.S}=\lim\limits_{\text{x}\rightarrow\infty}\big(\sqrt{\text{x}^2+\text{x}+1}-\text{x}\big)$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\big(\sqrt{\text{x}^2+\text{x}+1}-\text{x}\big)\big(\sqrt{\text{x}^2+\text{x}+1}+\text{x}\big)}{\big(\sqrt{\text{x}^2+\text{x}+1}+\text{x}\big)}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\big(\sqrt{\text{x}^2+\text{x}+1}-\text{x}^2\big)}{\sqrt{\text{x}^2+\text{x}+1}+\text{x}}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\text{x}\Big(1+\frac{1}{\text{x}}\Big)}{\text{x}\Big(1+\frac{1}{\text{x}}+\frac{1}{\text{x}^2}+1\Big)}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{1+\frac{1}{\text{x}}}{\sqrt{1+\frac{1}{\text{x}}+\frac{1}{\text{x}^2}}+1}$
$=\frac{1}{1+1}=\frac12$
$\therefore\ \lim\limits_{\text{x}\rightarrow\infty}\big(\sqrt{\text{x}^2+\text{x}+1}-\text{x}\big)$ is not equal to $\lim\limits_{\text{x}\rightarrow\infty}\big(\sqrt{\text{x}^2+1}-\text{x}\big).$
View full question & answer→Question 53 Marks
Evaluate the following limit:
$\lim\limits_{\text{n}\rightarrow\infty}\Big(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ \cdots+\frac{1}{3^\text{n}}\Big)$
Answer$\lim\limits_{\text{n}\rightarrow\infty}\Big(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ \cdots+\frac{1}{3^\text{n}}\Big)\ \cdots(\text{i})$
This is G.P. of common ratio $\frac13$
$\therefore$ Sum of n term of G.P. with $\text{a} = \frac13,\text{r}=\frac{-1}{3}$
$\text{S}_\text{n}=\text{a}\Big(\frac{1-\text{r}^{\text{n}}}{1-\text{r}}\Big)$
$\text{S}_\text{n}=\frac13\Bigg(\frac{1-\big(\frac{1}{3}\big)^\text{n}}{1-\frac13}\Bigg)$
$=\frac13\Bigg(\frac{1-\frac{1}{3^\text{n}}}{\frac23}\Bigg)$
$=\frac13\times\frac32\Big(1-\frac{1}{3^\text{n}}\Big)$
$\text{S}_\text{n}=\frac12\Big(1-\frac{1}{3^\text{n}}\Big)$
Substituting value of Sn in (i), we get
$\lim\limits_{\text{n}\rightarrow\infty}\text{S}_\text{n}=\lim\limits_{\text{n}\rightarrow\infty}\frac12\Big(1-\frac{1}{3^\text{n}}\Big)$
$=\frac{1}{2}\lim\limits_{\text{n}\rightarrow\infty}\Big(1-\frac{1}{3^\text{n}}\Big)$
$=\frac12(1-0)$
$=\frac12$
View full question & answer→Question 63 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow5}\frac{\text{x}^2-9\text{x}+20}{{\text{x}^2}-6\text{x}+5}$
Answer$\lim\limits_{\text{x}\rightarrow5}\frac{\text{x}^2-9\text{x}+20}{{\text{x}^2}-6\text{x}+5}$
$=\lim\limits_{\text{x}\rightarrow5}\frac{\text{x}^2-4\text{x}-5\text{x}+20}{\text{x}^2+\text{x}-5\text{x}+5}$
$=\lim\limits_{\text{x}\rightarrow5}\frac{\text{x}(\text{x}-4)-5(\text{x}-4)}{\text{x}(\text{x}-1)-5(\text{x}-1)}$
$=\lim\limits_{\text{x}\rightarrow5}\frac{(\text{x}-5)(\text{x}-4)}{(\text{x}-5)(\text{x}-1)}$
$=\lim\limits_{\text{x}\rightarrow5}\frac{\text{x}-4}{\text{x}-1}$
$=\frac{5-4}{5-1}$
$=\frac14$
View full question & answer→Question 73 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\cos\text{x}+2\sin\text{x}}{\text{x}^2+\tan\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\cos\text{x}+2\sin\text{x}}{\text{x}^2+\tan\text{x}}$
Dividing each term by x
$=\lim\limits_{\text{x}\rightarrow0}\frac{\cos\text{x}+\frac{2\sin\text{x}}{\text{x}}}{\text{x}+\frac{\tan\text{x}}{\text{x}}}$
$=\frac{\lim\limits_{\text{x}\rightarrow0}\cos\text{x}+\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}}{\text{x}}}{\lim\limits_{\text{x}\rightarrow0}\text{x}+\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}}$
$=\frac{\cos0+2\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}}{0+\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}}$
$=\frac{1+2}{0+1}=3$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1,\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$
$=3$
View full question & answer→Question 83 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\infty}}{\sqrt{\text{x}+1}-\sqrt{\text{x}}}{}$
Answer$\lim\limits_{\text{x}\rightarrow{\infty}}{\sqrt{\text{x}+1}-\sqrt{\text{x}}}{}$
$=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{\big(\sqrt{\text{x}+1}-\sqrt{\text{x}}\big)\big(\sqrt{\text{x}+1}+\sqrt{\text{x}}\big)}{\big(\sqrt{\text{x}+1}+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{\big(\text{x}+1-\text{x}\big)}{\sqrt{\text{x}+1}+\sqrt{\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow{\infty}}\bigg(\frac{1}{\sqrt{\text{x}+1}+\sqrt{\text{x}}}\bigg)$
$=\frac{1}{\infty}$
$=0$
View full question & answer→Question 93 Marks
Evaluate the following one sided limits:
$\lim\limits_{\text{x}\rightarrow2^-}\frac{\text{x}-3}{\text{x}^2-4}.$
Answer$\lim\limits_{\text{x}\rightarrow2^-}\frac{\text{x}-3}{(\text{x}^2-4)}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(2-\text{h})-3}{(2-\text{h)}^2-4^2}$ $\Big[\because\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2-\text{h)}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(2-\text{h}-3)}{(2-\text{h}+2)(2-\text{h}-2)}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-1-\text{h}}{(4-\text{h})(-\text{h})}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{1}{\text{h}}+1}{(4-\text{h})}$
$=\frac{\frac{1}{0}+1}{4}=\infty$
View full question & answer→Question 103 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow2}\Big(\frac{1}{\text{x}-2}-\frac{4}{\text{x}^3-2\text{x}}\Big)$
Answer$\lim\limits_{\text{x}\rightarrow2}\Big(\frac{1}{\text{x}-2}-\frac{4}{\text{x}^3-2\text{x}}\Big)$
$=\lim\limits_{\text{x}\rightarrow2}\Big(\frac{1}{\text{x}-2}-\frac{2}{\text{x}(\text{x}-2)}\Big)$
$=\lim\limits_{\text{x}\rightarrow2}\Big(\frac{(\text{x}-2)}{(\text{x}-2)(\text{x})}\Big)$
$=\lim\limits_{\text{x}\rightarrow2}\Big(\frac{(\text{x}-2)}{(\text{x}-2)(\text{x})}\Big)$
$=\lim\limits_{\text{x}\rightarrow2}\frac{1}{\text{x}}$
$=\frac12$
View full question & answer→Question 113 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{ax}+\text{x}\cos\text{x}}{\text{b}\sin\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{ax}+\text{x}\cos\text{x}}{\text{b}\sin\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}+\cos\text{x}}{\frac{\text{b}\sin\text{x}}{\text{x}}}$
$=\frac{\lim\limits_{\text{x} \rightarrow0}\text{a}+\lim\limits_{\text{x} \rightarrow0}\cos\text{x}}{\lim\limits_{\text{x}\rightarrow0}\frac{\text{b}\sin\text{x}}{\text{x}}}$
$=\frac{\text{a}+1}{\text{b}}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=\frac{\text{a}+1}{\text{b}}$
View full question & answer→Question 123 Marks
Evaluate the following limits:
$\lim\limits_{\text{x}\rightarrow\infty}\frac{3\text{x}^{-1}+4\text{x}^{-2}}{5\text{x}^{-1}+6\text{x}^{-2}}$
Answer$\lim\limits_{\text{x}\rightarrow\infty}\frac{3\text{x}^{-1}+4\text{x}^{-2}}{5\text{x}^{-1}+6\text{x}^{-2}}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\frac{3}{\text{x}}+\frac{4}{\text{x}^2}}{\frac{5}{\text{x}}+\frac{6}{\text{x}^2}}$ $\Big[\frac{0}{0}\text{ from}\Big]$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\frac{1}{\text{x}}\big(3+\frac{4}{\text{x}}\big)}{\frac{1}{\text{x}}\big(5+\frac{6}{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{(3+0)}{(5+0)}=\frac35$
View full question & answer→Question 133 Marks
Evaluate the following limit:
$\lim\limits_{\text{n}\rightarrow\infty}\frac{1^2+2^2+\ \dots+\text{n}^2}{\text{n}^4}$
Answer$\lim\limits_{\text{n}\rightarrow\infty}\frac{1^2+2^2+\ \dots+\text{n}^2}{\text{n}^4}$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{\Big[\frac12\text{n}(\text{n}+1)\Big]^2}{\text{n}^4}$ $\bigg[1^3+2^3+3^3+\ \cdots+\text{n}^3=\Big(\frac12\text{n}(\text{n}+1)\Big)^2\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{\frac14\text{n}^2(\text{n}+1)^2}{\text{n}^4}$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac14\frac{\big(\text{n}^2\big(\text{n}^2+1+2\text{n}\big)\big)}{\text{n}^4}$ $\Big[\text{Multiplying the term }\frac{\infty}{\infty}\text{ from}\Big]$
$=\frac{1}{4}\lim\limits_{\text{n}\rightarrow{\infty}}\frac{\Big(1+\frac{1}{\text{n}^2}+\frac{2}{\text{n}}\Big)}{1}$
$=\frac{1}{4}$
View full question & answer→Question 143 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos5\text{x}}{1-\cos6\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos5\text{x}}{1-\cos6\text{x}}$
$=\frac{\lim\limits_{\text{x}\rightarrow0}2\sin^2\frac{5\text{x}}{2}}{\lim\limits_{\text{x}\rightarrow0}2\sin^23\text{x}}$
$=\frac{2\Bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\frac{5\text{x}}{2}}{\frac{5\text{x}}{2}}\Bigg)^2\times\frac{25}{4}\text{x}^2}{2\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{3\text{x}}\Big)^2\times9\text{x}^2}$
$=\frac{2\times1\times\frac{25}{4}\text{x}^2}{2\times1\times9\text{x}^9}$
$=\frac{25}{4\times9}$
$=\frac{25}{36}$
View full question & answer→Question 153 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\cos\text{x}-\cos\text{a}}{\sqrt{\text{x}}-\sqrt{\text{a}}}$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\cos\text{x}-\cos\text{a}}{\sqrt{\text{x}}-\sqrt{\text{a}}}$
$=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\Big(-2\sin\big(\frac{\text{x}+\text{a}}{2}\big)\sin\big(\frac{\text{x}-\text{a}}{2}\big)\Big)\times\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)}{\big(\sqrt{\text{x}}-\sqrt{\text{a}}\big)\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)}$ $=-2\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\sin\big(\frac{\text{x}+\text{a}}{2}\big)\sin\big(\frac{\text{x}-\text{a}}{2}\big)\times\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)}{(\text{x}-\text{a})}$ $=-2\lim\limits_{\text{x}\rightarrow{\text{a}}}{\sin\big(\frac{\text{x}+\text{a}}{2}\big)}\times\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\sin\big(\frac{\text{x}-\text{a}}{2}\big)\times\frac12}{\big(\frac{\text{x}-\text{a}}{2}\big)}\lim\limits_{\text{x}\rightarrow{\text{a}}}\big(\sqrt{\text{x}}+\sqrt{\text{a}}\big)$ $=-2\times\sin(\text{a})\times1\times\frac12\times2\sqrt{\text{a}}$ $=-2\sqrt{\text{a}}\sin\text{a}$ View full question & answer→Question 163 Marks
Evaluate the following one sided limits:
$\lim\limits_{\text{x}\rightarrow-2^+}\frac{\text{x}^2-1}{2\text{x}+4}$
Answer $\lim\limits_{\text{x}\rightarrow-2^+}\frac{\text{x}^2-1}{2\text{x}+4}$
$=\lim\limits_{\text{x}\rightarrow2^+}\frac{(\text{x}-1)(\text{x}+1)}{2(\text{x}+2)}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(-2+\text{h}-1)(-2+\text{h}+1)}{2(-2+\text{h}+2)}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(-3+\text{h})(\text{h}-1)}{2\text{h}}$
$\Rightarrow\frac{-3\times-1}{2\times0}=\frac{1}{0}=\infty$
View full question & answer→Question 173 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}-\sin\text{x}}{\sin\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}-\sin\text{x}}{\sin\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\Bigg(\frac{2\cos\big(\frac{3\text{x}+\text{x}}{2}\big)\sin\big(\frac{3\text{x}-\text{x}}{2}\big)}{\sin\text{x}}\Bigg)$
$=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{2\cos2\text{x}\sin\text{x}}{\sin\text{x}}\Big)$
$=\lim\limits_{\text{x}\rightarrow0}(2\cos\text{x})$
$=2\lim\limits_{\text{x}\rightarrow0}\cos2\text{x}$
$=2\times\cos0$
$=2\times1=2$
$=2$
View full question & answer→Question 183 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{2\text{x}}{\sqrt{\text{a}+\text{x}}-\sqrt{\text{a}-\text{x}}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{2\text{x}}{\sqrt{\text{a}+\text{x}}-\sqrt{\text{a}-\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\text{x}}{\big(\sqrt{\text{a}+\text{x}}-\sqrt{\text{a}-\text{x}}\big)}\times\frac{\sqrt{\text{a}+\text{x}}+\sqrt{\text{a}-\text{x}}}{\big(\sqrt{\text{a}+\text{x}}+\sqrt{\text{a}-\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\text{x}\big(\sqrt{\text{a}+\text{x}}+\sqrt{\text{a}-\text{x}}\big)}{\big(\big(\text{a}+\text{x}-\big(\text{a}-\text{x}\big)\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\text{x}\big(\sqrt{\text{a}+\text{x}}+\sqrt{\text{a}-\text{x}}\big)}{2\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\big(\sqrt{\text{a}+\text{x}}+\sqrt{\text{a}-\text{x}}\big)$
$=\sqrt{\text{a}}+\sqrt{\text{a}}$
$=2\sqrt{\text{a}}$
View full question & answer→Question 193 Marks
Find $\lim\limits_{\text{x}\rightarrow3^+}\frac{\text{x}}{[\text{x}]}.$ is it equal to $\lim\limits_{\text{x}\rightarrow3^-}\frac{\text{x}}{[\text{x}]}.$
Answer$\lim\limits_{\text{x}\rightarrow3^+}\frac{\text{x}}{[\text{x}]}=\lim\limits_{\text{x}\rightarrow3^+}\frac{\text{x}}{3}=\frac{3}{3}=1$
$\lim\limits_{\text{x}\rightarrow3^-}\frac{\text{x}}{[\text{x}]}=\lim\limits_{\text{x}\rightarrow3^-}\frac{\text{x}}{2}=\frac{3}{2}=1.5$
$\therefore\ \lim\limits_{\text{x}\rightarrow3^+}\frac{\text{x}}{[\text{x}]}\ne\lim\limits_{\text{x}\rightarrow3^-}\frac{\text{x}}{[\text{x}]}$
View full question & answer→Question 203 Marks
Evaluate the following limit:
$\lim\limits_{\theta\rightarrow0}\frac{\sin3\theta}{\tan2\theta}$
Answer$\lim\limits_{\theta\rightarrow0}\frac{\sin3\theta}{\tan2\theta}$ $=\frac{\lim\limits_{\theta\rightarrow0}\sin3\theta}{\lim\limits_{\theta\rightarrow0}\tan2\theta}$ $=\frac{\lim\limits_{3\theta\rightarrow0}\frac{\sin3\theta}{3\theta}\times3\theta}{\lim\limits_{2\theta\rightarrow0}\frac{\tan2\theta}{2\theta}\times2\theta}$ $=\frac{\Big(\lim\limits_{3\theta\rightarrow0}\frac{\sin3\theta}{3\theta}\Big)}{\Big(\lim\limits_{2\theta\rightarrow0}\frac{\tan2\theta}{2\theta}\Big)}\times\frac{3\theta}{2\theta}$ $=\frac11\times\frac32$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\text{ and }\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$
$=\frac32$ View full question & answer→Question 213 Marks
Evaluate the following one sided limits:
$\lim\limits_{\text{x}\rightarrow0^-}(2-\cos\text{x})$
Answer $\lim\limits_{\text{x}\rightarrow0^-}(2-\cos\text{x})$
$\lim\limits_{\text{h}\rightarrow0}\ 2-\cot(0-\text{h})$
$=\lim\limits_{\text{h}\rightarrow0}\ 2-(-1)\cot\text{ h}$
$=\lim\limits_{\text{h}\rightarrow0}2+\cot\text{ h}$
$=\lim\limits_{\text{h}\rightarrow0}\ 2+\frac{1}{\tan\text{h}}$
$\Rightarrow2+\frac{1}{0}\Leftarrow\infty$
View full question & answer→Question 223 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\text{f(x)}-\text{f}\big(\frac\pi4\big)}{\text{x}-\frac\pi4},$ where $\text{f(x)}=\sin2\text{x}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\Bigg(\frac{\sin2\text{x}-\sin2\big(\frac\pi4\big)}{\text{x}-\frac\pi4}\Bigg)$ $\big[\because$ given $\text{f(x)}=\sin2\text{x}\big]$
$=\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\bigg(\frac{\sin2\text{x}-\sin\frac{\pi}{2}}{\text{x}-\frac\pi4}\bigg)$
$\Rightarrow\text{x}\rightarrow\frac{\pi}{4}\Rightarrow\text{x}-\frac{\pi}{4}\rightarrow0,$ let $\text{x}-\frac\pi4=\text{y}$
$=\lim\limits_{\text{y}\rightarrow{0}}\Bigg(\frac{\sin2\big(\text{y}+\frac{\pi}{4}\big)-1}{\text{y}}\Bigg)$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\big(\frac\pi2+2\text{y}\big)-1}{\text{y}}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\cos2\text{y}-1}{\text{y}}$
$=-\lim\limits_{\text{y}\rightarrow{0}}\frac{1-\cos2\text{y}}{\text{y}}$
$=-\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\text{y}}{\text{y}}$
$=-2\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\text{y}}{\text{y}}\Big)^2\times\text{y}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=-2\times0$
$=0$
View full question & answer→Question 233 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}(\text{cosec x}-\cot\text{x})$
Answer$\lim\limits_{\text{x}\rightarrow0}(\text{cosec x}-\cot\text{x})$
$=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}\Big)$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin^2\frac{\text{x}}{2}}{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{2}$
$=\bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\tan\text{x}}{2}}{\frac{\text{x}}{2}}\bigg)\times\frac{\text{x}}{2}$
$=\lim\limits_{\text{x}\rightarrow0}1\times\frac{\text{x}}{2}$
$=0$
View full question & answer→Question 243 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}(\cos\text{x})^{\frac{1}{\sin\text{x}}}$
Answer$\lim\limits_{\text{x}\rightarrow0}(\cos\text{x})^{\frac{1}{\sin\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow0}(1+\cos\text{x}-1)^{\frac{1}{\sin\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow0}(1-(1-\cos\text{x}))^{\frac{1}{\sin\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow0}\Big(1-2\sin^2\Big(\frac{\text{x}}{2}\Big)\Big)^{\frac{1}{\sin\text{x}}}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow0}\Big(-2\sin^2\Big(\frac{\text{x}}{2}\Big)\Big)\times\Big(\frac{1}{\sin\text{x}}\Big)}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow0}\Bigg(\frac{-2\sin^2\big(\frac{\text{x}}{2}\big)}{\sin\text{x}}\Bigg)}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow0}\begin{pmatrix}\frac{-2\sin^2\big(\frac{\text{x}}{2}\big)}{2\sin\big(\frac{\text{x}}{2}\big)\cos\big(\frac{\text{x}}{2}\big)}\end{pmatrix}}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow0}-\tan\text{x}}$
$=\ \text{e}^0$
$=\ 1$
View full question & answer→Question 253 Marks
Find $\lim\limits_{\text{x}\rightarrow3}\text{f(x)},$ where $\text{f(x)}=\begin{cases}4, & \text{if x}> 3\\\text{x}+1, &\text{if x} < 3\end{cases}.$
Answer$\lim\limits_{\text{x}\rightarrow3^+}\text{f(x)}=4$
$\lim\limits_{\text{x}\rightarrow3^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow3^-}\text{(x}+1)=\lim\limits_{\text{h}\rightarrow0}(3-\text{h}+1)=3+1=4$
Since, $\lim\limits_{\text{x}\rightarrow3^+}\text{f(x)}=4=\lim\limits_{\text{x}\rightarrow3^-}\text{f(x)}$
$\therefore\ \lim\limits_{\text{x}\rightarrow3}\text{f(x) is }4$
View full question & answer→Question 263 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow3}\big(\text{x}^2-9\big)\Big(\frac{1}{\text{x}+3}-\frac{1}{\text{x}-3}\Big)$
Answer$\lim\limits_{\text{x}\rightarrow3}\big(\text{x}^2-9\big)\Big(\frac{1}{\text{x}+3}-\frac{1}{\text{x}-3}\Big)$
$=\lim\limits_{\text{x}\rightarrow3}\big(\text{x}^2-9\big)\Big(\frac{\text{x}-3+\text{x}+3}{(\text{x}+3)(\text{x}-3)}\Big)$
$=\lim\limits_{\text{x}\rightarrow3}\big(\text{x}^2-9\big)\Big(\frac{2\text{x}}{(\text{x}+3)(\text{x}-3)}\Big)$
$=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}-3)(\text{x}+3)(2\text{x})}{(\text{x}+3)(\text{x}-3)}$
$=\lim\limits_{\text{x}\rightarrow3}2(\text{x})=2(3)=6$
View full question & answer→Question 273 Marks
If $\text{f(x)}=\begin{cases}2\text{x}+3, & \text{x} \le0\\3(\text{x}+1), &\text{x} > 0\end{cases}$Find $\lim\limits_{\text{x}\rightarrow0}\text{f(x)}$ and $\lim\limits_{\text{x}\rightarrow1}\text{f(x)}.$
Answer$\lim\limits_{\text{x}\rightarrow0^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0^+}3\text{(x}+1)=3$
And,
$\lim\limits_{\text{x}\rightarrow0^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0^-}2\text{(x)}+3=3$
$\therefore\ \lim\limits_{\text{x}\rightarrow0}\text{f(x)}=3$
$\lim\limits_{\text{x}\rightarrow1}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1}3(\text{x}+1)=3+3=6$
$\therefore\ \lim\limits_{\text{x}\rightarrow1}\text{f(x)}=6$
View full question & answer→Question 283 Marks
Let $\text{f(x)}=\begin{cases}\text{x}+5, & \text{if x}> 0\\\text{x}-4, &\text{if x} < 0\end{cases}.$ Prove that $\lim\limits_{\text{x}\rightarrow0}\text{f(x)}$ does not exist.
Answer$\text{L.H.L}=\lim\limits_{\text{x}\rightarrow0^-}\text{f(x)}$
$=\lim\limits_{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$\lim\limits_{\text{h}\rightarrow0}-\text{h}-4$
$=0-4$
$=-4$
$\text{R.H.L}=\lim\limits_{\text{x}\rightarrow0^+}\text{f(x)}$
$=\lim\limits_{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}+5$
$=0+5$
$=5$
$\therefore\ \lim\limits_{\text{x}\rightarrow0^-}\text{f(x)}\neq\lim\limits_{\text{x}\rightarrow0^+}\text{f(x)}$
Hence $\lim\limits_{\text{x}\rightarrow0}\text{f(x)}$ does not exist.
View full question & answer→Question 293 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow3}\frac{\sqrt{\text{x}+3}-\sqrt{6}}{\text{x}^2-9}$
Answer$\lim\limits_{\text{x}\rightarrow3}\frac{\sqrt{\text{x}+3}-\sqrt{6}}{\text{x}^2-9}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{\big(\sqrt{\text{x}+3}-\sqrt{6}\big)\big(\sqrt{\text{x}+3}-\sqrt{6}\big)}{(\text{x}-3)(\text{x}+3)\big(\sqrt{\text{x}+3}+\sqrt{6}\big)}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{((\text{x}+3)-6)}{(\text{x}-3)(\text{x}+3)\big(\sqrt{\text{x}+3}+\sqrt{6}\big)}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}-3)}{(\text{x}-3)(\text{x}+3)\big(\sqrt{\text{x}+3}+\sqrt{6}\big)}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{1}{(\text{x}+3)\big(\sqrt{\text{x}+3}+\sqrt{6}\big)}$
$=\frac{1}{(3+3)\sqrt{3+3}+\sqrt{6}}$
$=\frac{1}{6\big(\sqrt{6}+\sqrt{6}\big)}=\frac{1}{6\times2\sqrt{6}}$
$=\frac{1}{12\sqrt{6}}$
View full question & answer→Question 303 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin(2+\text{x})-\sin(2-\text{x})}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin(2+\text{x})-\sin(2-\text{x})}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}2\cos\frac{\Big(\frac{{2+\text{x}+2-\text{x}}{2}}{\text{x}}\Big)\times\sin\Big(\frac{2+\text{x}-2+\text{x}}{2}\Big)}{{\text{x}}}$
$=2\lim\limits_{\text{x}\rightarrow0}\frac{\cos(2)\times\sin\text{x}}{\text{x}}$
$=2\lim\limits_{\text{x}\rightarrow0}\cos2\times\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=2\cos2\times1$
$=2\cos2$
View full question & answer→Question 313 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sec5\text{x}-\sec3\text{x}}{\sec3\text{x}-\sec\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}-\sin\text{x}}{\sin3\text{x}-3\sin\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\Bigg(\frac{\frac{\cos3\text{x}-\cos5\text{x}}{\cos3\text{x}\cos5\text{x}}}{\frac{\cos\text{x}-\cos3\text{x}}{\cos\text{x}\cos3\text{x}}}\Bigg)$
$=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\cos3\text{x}-\cos5\text{x}}{\cos\text{x}-\cos3\text{x}}\times\frac{\cos\text{x}\cos3\text{x}}{\cos3\text{x}\cos5\text{x}}\Big)$
$=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{-2\sin4\text{x}\sin(-\text{x})}{-2\sin(2\text{x})\sin(-\text{x})}\times\frac{\cos\text{x}}{\cos5\text{x}}\Big)$
$=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\sin4\text{x}}{\sin2\text{x}}\times\frac{\cos\text{x}}{\cos5\text{x}}\Big)$
$=\frac{\lim\limits_{\text{x}\rightarrow0}\sin4\text{x}\times\lim\limits_{\text{x}\rightarrow0}\cos\text{x}}{\lim\limits_{\text{x}\rightarrow0}\sin2\text{x}\times\lim\limits_{\text{x}\rightarrow0}\cos5\text{x}}$
$=\frac{\Big(\lim\limits_{4\text{x}\rightarrow0}\frac{\sin4\text{x}}{4\text{x}}\times4\text{x}\Big)\big(\lim\limits_{\text{x}\rightarrow0}\cos\text{x}\big)}{\Big(\lim\limits_{2\text{x}\rightarrow0}\frac{\sin2\text{x}}{2}\times2\text{x}\Big)\big(\lim\limits_{\text{x}\rightarrow0}\cos5\text{x}\big)}$
$=\frac{(1\times4\text{x})\times1}{1\times2\text{x}\times1}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\text{ and }\lim\limits_{\text{x}\rightarrow0}\cos\text{x}=\cos0=1\Big]$
$=\frac{4\text{x}}{2\text{x}}$
$=2$
View full question & answer→Question 323 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\sqrt{3}}\frac{\text{x}^4-9}{{\text{x}^2+4\sqrt{3}\text{x}-15}}$
Answer$\lim\limits_{\text{x}\rightarrow\sqrt{3}}\frac{\text{x}^4-9}{{\text{x}^2+4\sqrt{3}\text{x}-15}}$
$=\lim\limits_{\text{x}\rightarrow\sqrt{3}}\frac{\big(\text{x}-\sqrt{3}\big)\big(\text{x}+\sqrt{3}\big)\big(\text{x}^2+3\big)}{{\big(\text{x}-\sqrt{3}\big)\big(\text{x}+5\sqrt{3}\big)}}$
$=\lim\limits_{\text{x}\rightarrow\sqrt{3}}\frac{\big(\text{x}+\sqrt{3}\big)\big(\text{x}^2+3\big)}{{\big(\text{x}+5\sqrt{3}\big)}}$
$=\frac{\big(\sqrt{3}+\sqrt{3}\big)(3+3)}{\big(\sqrt{3}+5\sqrt{3}\big)}=\frac{\big(3\sqrt{3}\big)(6)}{6\sqrt{3}}=2$
View full question & answer→Question 333 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin^24\text{x}^2}{\text{x}^4}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin^24\text{x}^2}{\text{x}^4}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sin4{\text{x}^2\big)}^{2}}{\text{x}^4}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sin4\text{x}^2\big)^2}{\big(\text{x}^2\big)^2}$ $=\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin4\text{x}^2}{\text{x}^2}\Big)^2$ $=\Big(\lim\limits_{4\text{x}^2\rightarrow0}\frac{\sin4\text{x}^2}{4\text{x}^2}\Big)\times16$ $=1\times16$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$ $=16$
View full question & answer→Question 343 Marks
Evaluate the following limit:
$\lim\limits_{\theta\rightarrow0}\frac{\sin4\theta}{\tan3\theta}$
Answer$\lim\limits_{\theta\rightarrow0}\frac{\sin4\theta}{\tan3\theta}$
$=\frac{\lim\limits_{\theta\rightarrow0}\sin4\theta}{\lim\limits_{\theta\rightarrow0}\tan3\theta}$
$=\frac{\Big(\lim\limits_{\theta\rightarrow0}\frac{\sin4\theta}{4\theta}\Big)\times4\theta}{\Big(\lim\limits_{\theta\rightarrow0}\frac{\tan3\theta}{3\theta}\Big)\times3\theta}$
$=\frac{1\times4\theta}{1\times3\theta}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=\frac43$
View full question & answer→Question 353 Marks
Find k so that $\lim\limits_{\text{x}\rightarrow2}\text{f(x)}$ may exist, where $\text{f(x)}=\begin{cases}2\text{x}+3, & \text{x}\le 2\\\text{x}+\text{k}, & \text{x} > 2\end{cases}.$
Answer$\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow2^-}(2\text{x}+3)$
$=2(2)+3$
$=7$
$\therefore\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=7$
Also,
$\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow2^+}(\text{x}+\text{k})$
$=(2+\text{k})$
Since, $\lim\limits_{\text{x}\rightarrow2}\text{f(x)}$ exists (given)
$\therefore\ \lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}$
$\Rightarrow7=2+\text{k}$
$\Rightarrow\text{k}=5$
View full question & answer→Question 363 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin5\text{x}-\sin3\text{x}}{\sin\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin5\text{x}-\sin3\text{x}}{\sin\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\cos\big(\frac{5\text{x}+3\text{x}}{2}\big)\sin\big(\frac{5\text{x}-3\text{x}}{2}\big)}{\sin\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\cos4\text{x}\sin\text{x}}{\sin\text{x}}$
$=2\lim\limits_{\text{x}\rightarrow0}\cos4\text{x}$
$=2\times\cos0$
$=2$
View full question & answer→Question 373 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\sqrt{2}-\cos\text{x}-\sin\text{x}}{(4\text{x}-\pi)^2}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\sqrt{2}-\cos\text{x}-\sin\text{x}}{(4\text{x}-\pi)^2}$
$\text{x}\rightarrow\frac{\pi}{4}$ then $\text{x}-\frac\pi4\rightarrow0,$ also $4\text{x}-\pi\rightarrow0$ let $\text{x}-\frac{\pi}{4}\rightarrow\text{y}$
$=\lim\limits_{\text{x}-\frac{\pi}{4}\rightarrow{0}}\frac{\sqrt{2}-\cos\text{x}-\sin\text{x}}{(4)^2\big(\text{x}-\frac{\pi}{4}\big)^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\cos\big({\text{y}+\frac\pi4\big)-\sin\big({\text{y}+\frac\pi4\big)}}}{16\times\text{y}^2}$
$=\frac{1}{16}\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\big(\cos\text{y}\times\frac{1}{\sqrt{2}}-\sin\text{y}\times\frac{1}{\sqrt{2}}\big)-\big(\frac{\sin\text{y}}{\sqrt{2}}+\frac{\cos\text{y}}{\sqrt{2}}\big)}{\text{y}^2}$
$=\frac{1}{16}\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\frac{1}{\sqrt{2}}(\cos\text{y}-\sin\text{y})-\frac{1}{\sqrt{2}}(\sin\text{y}+\cos\text{y})}{\text{y}^2}$
$=\frac{1}{16}\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\frac{1}{\sqrt{2}}\big[(\cos\text{y}-\sin\text{y})-(\sin\text{y}+\cos\text{y})\big]}{\text{y}^2}$
$=\frac{1}{16}\lim\limits_{\text{y}\rightarrow{0}}\frac{\Big(\sqrt{2}-\frac{1}{\sqrt{2}}\cos\text{y}+\frac{1}{\sqrt{2}}\sin\text{y}-\frac{1}{\sqrt{2}}\sin\text{y}-\frac{1}{\sqrt{2}}\cos\text{y}\Big)}{\text{y}^2}$
$=\frac{1}{16}\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\frac{1}{\sqrt{2}}\cos\text{y}}{\text{y}^2}$
$=\frac{\sqrt{2}}{16}\lim\limits_{\text{y}\rightarrow{0}}\frac{{2}\sin^2\frac{\text{y}}{{2}}}{\text{y}^2}$
$=\frac{\sqrt{2}}{8}\lim\limits_{\text{y}\rightarrow{0}}\Bigg(\frac{\sin\frac{\text{y}}{{2}}}{\frac{\text{y}}{2}}\Bigg)^2\times\frac14$
$=\frac{1}{16\sqrt{2}}$
View full question & answer→Question 383 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow2}\Big(\frac{\text{x}}{\text{x}-3}-\frac{4}{\text{x}^2-2\text{x}}\Big)$
Answer$\lim\limits_{\text{x}\rightarrow2}\Big(\frac{\text{x}}{\text{x}-2}-\frac{4}{\text{x}^2-2\text{x}}\Big)$
$=\lim\limits_{\text{x}\rightarrow2}\Big(\frac{\text{x}}{\text{x}-2}-\frac{4}{\text{x}(\text{x}-2)}\Big)$
$=\lim\limits_{\text{x}\rightarrow2}\Big(\frac{\text{x}(\text{x}-4)}{\text{x}(\text{x}-2)}\Big)$
$=\lim\limits_{\text{x}\rightarrow2}\Big(\frac{\text{x}^2-4}{\text{x}(\text{x}-2)}\Big)$
$=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}-2)(\text{x}+2)}{\text{x}(\text{x}-2)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}+2)}{\text{x}}$
$=\frac{2+2}{2}$
$=\frac42$
$=2$
View full question & answer→Question 393 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^3\cot\text{x}}{1-\cos\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^3\cot\text{x}}{1-\cos\text{x}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}^3}{\tan\text{x}(1-\cos\text{x})}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}^3}{\tan\text{x}.2\sin^2\frac{\text{x}}{2}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{1}{\frac{\tan\text{x}}{\text{x}}\times\frac{2\sin^2\frac{\text{x}}{2}}{\text{x}^2}}$
$=\frac{1}{\bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\tan\text{x}}{\text{x}}\bigg)\times2\bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\bigg)^2\times\frac14}$
$=\frac{1}{2\times2\times1\times\frac14}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1,\lim\limits_{\theta\rightarrow0}\frac{\tan\theta}{\theta}=1\Big]$
$=2$
View full question & answer→Question 403 Marks
Show that $\lim\limits_{\text{x}\rightarrow2^-}\ \frac{\text{x}}{[\text{x}]}\ne\lim\limits_{\text{x}\rightarrow2^+}\frac{\text{x}}{[\text{a}]}.$
Answer $\lim\limits_{\text{x}\rightarrow2^-}\ \frac{\text{x}}{[\text{x}]}$
$=\lim\limits_{\text{x}\rightarrow2^-}\ \frac{\text{x}}{1}=\frac{2}{1}=2$ $\bigg[\because\lim\limits_{\text{x}\rightarrow\text{k}^-}\ [\text{x}]=\text{k}-1\bigg]$
Also,
$\lim\limits_{\text{x}\rightarrow2^+}\frac{\text{x}}{[\text{x}]}=\lim\limits_{\text{x}\rightarrow2^-}\frac{\text{x}}{3}=\frac23$ $\bigg[\because\lim\limits_{\text{x}\rightarrow\text{k}^+}\ [\text{x}]=\text{k}+1\bigg]$
$\Rightarrow\lim\limits_{\text{x}\rightarrow2^-}\ \frac{\text{x}}{[\text{x}]}\ne\lim\limits_{\text{x}\rightarrow2^+}\frac{\text{x}}{[\text{x}]}$
View full question & answer→Question 413 Marks
Let $\text{f(x)}=\begin{cases}\frac{\text{k}\cos\text{x}}{\pi-2\text{x}}, & \text{where x} \ne\frac\pi2\\3, & \text{where x} \ne\frac\pi2\end{cases}$ and if $\lim\limits_{\text{x}\rightarrow\frac\pi2}{\text{f(x)}}=\text{f}\Big(\frac\pi2\Big),$ find the value of k.
AnswerWe have,
$\text{f(x)}=\begin{cases}\frac{\text{k}\cos\text{x}}{\pi-2\text{x}}, & \text{where x} \ne\frac\pi2\\3, & \text{where x} \ne\frac\pi2\end{cases}$
It is given that,
$\lim\limits_{\text{x}\rightarrow\frac\pi2}{\text{f(x)}}={{\text{f}\big(\frac{\pi}{2}\big)}}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow\frac\pi2}\ \frac{{\text{k}\cos{\text{x}}}}{\pi-2\text{x}}=3$
$\Rightarrow\frac{{\text{k}}}{2}\times\lim\limits_{\text{x}-\frac\pi2\rightarrow0}\ \frac{\sin\big(\frac{\pi}{2}-{\text{x}}\big)}{\frac\pi2-{\text{x}}}=3$
$\Rightarrow\frac{\text{k}}{2}\times\lim\limits_{\text{h}\rightarrow0}\ \frac{\sin(-\text{h)}}{-\text{h}}=3$ $\big(\text{Put }\text{x}-\frac\pi2=\text{h}\big)$
$\Rightarrow\frac{\text{k}}{2}\times\lim\limits_{\text{h}\rightarrow0}\ \frac{\sin\text{h}}{\text{h}}=3$ $[\sin(-\theta)=-\sin\theta]$
$\Rightarrow\frac{\text{k}}{2}=3$ $\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big)$
$\Rightarrow\text{k}=6$
Hence, the value of k is 6.
View full question & answer→Question 423 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\cos3\text{x}-\cos7\text{x}}{\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\cos3\text{x}-\cos7\text{x}}{\text{x}^2}$
$=\lim\limits_{\text{x}\rightarrow0}\Bigg(\frac{-2\sin\big(\frac{3\text{x}+7\text{x}}{2}\big)\sin\big(\frac{3\text{x}-7\text{x}}{2}\big)}{\text{x}^2}\Bigg)$
$=\lim\limits_{\text{x}\rightarrow0}\Bigg(\frac{-2\sin5\text{x}\sin\big(\frac{-4\text{x}}{2}\big)}{\text{x}^2}\Bigg)$
$=\Big(\lim\limits_{\text{x}\rightarrow0}\frac{-2\sin5\text{x}}{\text{x}}\Big)\times\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin(-2\text{x})}{\text{x}}\Big)$
$=\Big(-2\lim\limits_{\text{x}\rightarrow0}\frac{\sin5\text{x}}{5\text{x}}\Big)\times\Big(-1\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}}{2\text{x}}\times2\Big)$
$=(-2\times5)(-1\times2)$
$=20$
View full question & answer→Question 433 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{5\text{x}\cos\text{x}+\sin\text{x}}{3\text{x}^2+\tan\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{5\text{x}\cos\text{x}+\sin\text{x}}{3\text{x}^2+\tan\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{5\cos\text{x}+3\frac{\sin\text{x}}{\text{x}}}{3\text{x}+\frac{\tan\text{x}}{\text{x}}}$
$=\frac{\lim\limits_{\text{x}\rightarrow0}5\cos\text{x}\lim\limits_{\text{x}\rightarrow0}\frac{3\sin\text{x}}{\text{x}}}{\lim\limits_{\text{x}\rightarrow0}3\text{x}+\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}}$
$=\frac{\lim\limits_{\text{x}\rightarrow0}5\cos\text{x}+3\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}}{3\lim\limits_{\text{x}\rightarrow0}\text{x}+\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}}$
$=\frac{5\times\cos0+3\times1}{3\times0+1}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1,\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$
$=\frac{5+3}{1}$
$=8$
View full question & answer→Question 443 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow3}\Big(\frac{1}{\text{x}-3}-\frac{2}{\text{x}^2-4\text{x}+3}\Big)$
Answer $\lim\limits_{\text{x}\rightarrow3}\Big(\frac{1}{\text{x}-3}-\frac{2}{\text{x}^2-4\text{x}+3}\Big)$
$=\lim\limits_{\text{x}\rightarrow3}\Big(\frac{1}{\text{x}-3}-\frac{2}{(\text{x}-3)(\text{x}-1)}\Big)$
$=\lim\limits_{\text{x}\rightarrow3}\Big(\frac{\text{x}-1-2}{(\text{x}-1)(\text{x}-3)}\Big)$
$=\lim\limits_{\text{x}\rightarrow3}\bigg(\frac{\text{x}-3}{(\text{x}-1)(\text{x}-3)}\bigg)$
$=\lim\limits_{\text{x}\rightarrow3}\frac{1}{\text{x}-1}$
$=\frac{1}{3-1}$
$=\frac12$
View full question & answer→Question 453 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{1-\sin\text{x}}{\big(\frac{\pi}{2}-\text{x}\big)}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{1-\sin\text{x}}{\big(\frac{\pi}{2}-\text{x}\big)}$
If $\text{x}\rightarrow\frac{\pi}{2},\frac{\pi}{2}-\text{x}\rightarrow0$
Let $\frac\pi2-\text{x}=\text{y}$ they y → 0
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{1-\sin\big(\frac{\pi}{4}-\text{y}\big)}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{1-\cos\text{y}}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\frac{\text{y}}{2}}{\text{y}^2}$
$=2\Bigg(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\frac{\text{y}}{2}}{\frac{\text{y}}{2}}\Bigg)^2\times\frac14$ $\Big[\because\lim\limits_{\text{x}\rightarrow{0}}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=2\times1\times\frac14$
$=\frac12$
View full question & answer→Question 463 Marks
Let a1, a2, ..., an be fixed real numbers such that f(x) = (x - a1)(x - a2) ...(x - an) What is $\lim\limits_{\text{x}\rightarrow\text{a}_1}\text{f(x)}?$ For $\text{a}\ne\text{a}_1,\text{a}_2,\dots\text{a}_\text{n}$ compute $\lim\limits_{\text{x}\rightarrow{\text{a}}}\text{f(x)}.$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\text{f(x)}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow\text{a}_1}({\text{x}-{\text{a}_1}})(\text{x}-\text{a}_2)\dots(\text{x}-\text{a}_\text{n})$ [Putting limits x → a1]
$\Rightarrow(\text{a}_1-\text{a}_1)(\text{a}_1-\text{a}_2)\dots(\text{a}_1-\text{a}_\text{n})$
$\Rightarrow0$
And,
$\lim\limits_{\text{x}\rightarrow{\text{a}}}\text{f(x)}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow{\text{a}}}(\text{x}-\text{a}_1)(\text{x}-\text{a}_2)\dots(\text{x}-\text{a}_\text{n})$ [Putting limit x → a]
$\Rightarrow(\text{a}-\text{a}_1)(\text{a}-\text{a}_2)\dots(\text{a}-\text{a}_\text{n}).$
View full question & answer→Question 473 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\sqrt{3}}\frac{\text{x}^2-3}{{\text{x}^2+3\sqrt{3}\text{x}-12}}$
Answer$\lim\limits_{\text{x}\rightarrow\sqrt{3}}\frac{\text{x}^2-3}{{\text{x}^2+3\sqrt{3}\text{x}-12}}$
$=\lim\limits_{\text{x}\rightarrow\sqrt{3}}\frac{\big(\text{x}-\sqrt{3}\big)\big(\text{x}+\sqrt{3}\big)}{\text{x}^2+4\sqrt{3\text{x}}-\sqrt{3\text{x}}-12}$
$=\lim\limits_{\text{x}\rightarrow\sqrt{3}}\frac{\big(\text{x}-\sqrt{3}\big)\big(\text{x}+\sqrt{3}\big)}{\text{x}\big(\text{x}+4\sqrt{3}\big)-\sqrt{3}\big(\text{x}+4\sqrt{3}\big)}$
$=\lim\limits_{\text{x}\rightarrow\sqrt{3}}\frac{\big(\text{x}-\sqrt{3}\big)\big(\text{x}+\sqrt{3}\big)}{\big(\text{x}-\sqrt{3}\big)\big(\text{x}+4\sqrt{3}\big)}$
$=\frac{\sqrt{3}+\sqrt{3}}{\sqrt{3}+4\sqrt{3}}=\frac{2\sqrt{3}}{5\sqrt{3}}$
$=\frac25$
View full question & answer→Question 483 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow1}\frac{1-\text{x}^{\frac{-1}{3}}}{1-\text{x}^{\frac{-2}{3}}}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{1-\text{x}^{\frac{-1}{3}}}{1-\text{x}^{\frac{-2}{3}}}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{1-\frac{1}{\text{x}^{\frac{1}{3}}}}{{1-\frac{1}{\text{x}^{\frac23}}}}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\Big(\text{x}^{\frac13}-1\Big)}{\Big(\text{x}^{\frac{1}{3}}-1\Big)\Big(\text{x}^{\frac13}+1\Big)}\times\text{x}^{\frac{1}{3}}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^{\frac{1}{3}}}{\text{x}^{\frac{1}{3}}+1}$
$=\frac{1}{1+1}$
$=\frac{1}{2}$
View full question & answer→Question 493 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\sqrt{2}}\frac{\text{x}^2-2}{{\text{x}^2+\sqrt{2}\text{x}-4}}$
Answer$\lim\limits_{\text{x}\rightarrow\sqrt{2}}\frac{\text{x}^2-2}{{\text{x}^2+\sqrt{2}\text{x}-4}}$
$=\lim\limits_{\text{x}\rightarrow\sqrt{2}}\frac{\big(\text{x}-\sqrt{2}\big)\big(\text{x}+\sqrt{2}\big)}{\text{x}^2+2\sqrt{2}\text{x}-\sqrt{2}\text{x}-4}$
$=\lim\limits_{\text{x}\rightarrow\sqrt{2}}\frac{\big(\text{x}-\sqrt{2}\big)\big(\text{x}+\sqrt{2}\big)}{\text{x}\big(\text{x}+2\sqrt{2}\big)-\sqrt{2\big(\text{x}+2\sqrt{2}\big)}}$
$=\lim\limits_{\text{x}\rightarrow\sqrt{2}}\frac{\big(\text{x}-\sqrt{2}\big)\big(\text{x}+\sqrt{2}\big)}{\big(\text{x}+2\sqrt{2}\big)\big(\text{x}-\sqrt{2}\big)}$
$=\frac{\sqrt{2}+\sqrt{2}}{\sqrt{2}+2\sqrt{2}}=\frac{2\sqrt{2}}{3\sqrt{2}}$
$=\frac23$
View full question & answer→Question 503 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{1-\tan\text{x}}{1-\sqrt{2}\sin\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{1-\tan\text{x}}{1-\sqrt{2}\sin\text{x}}$
$\text{x}\rightarrow\frac{\pi}{4},$ then $\text{x}-\frac{\pi}{4}\rightarrow0,$ let $\text{x}-\frac\pi4=\text{y}$
$\Rightarrow\lim\limits_{{\text{x}\rightarrow{\frac{\pi}{4}}}}\frac{1-\tan\text{x}}{1-\sqrt{2}\sin\text{x}}=\lim\limits_{{\text{x}-{\frac{\pi}{4}}}\rightarrow0}\frac{1-\tan\text{x}}{1-\sqrt{2}\sin\text{x}}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{1-\tan\big(\text{y}+\frac{\pi}{4}\big)}{1-\sqrt{2}\sin\big(\text{y}+\frac\pi4\big)}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{1-\Bigg(\frac{\tan\frac\pi4+\tan\text{y}}{1+\tan\frac\pi4+\tan\text{y}}\Bigg)}{1-\sqrt{2}\Big(\sin\text{y}\cos\frac{\pi}{4}+\cos\text{y}\sin\frac\pi4\Big)}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{\Big(1-\Big(\frac{1+\tan\text{y}}{1-\tan\text{y}}\Big)\Big)}{1-\sqrt{2}\Big(\frac{\sin\text{y}}{\sqrt{2}}+\frac{\cos\text{y}}{\sqrt{2}}\Big)}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{(1-\tan\text{y}-1-\tan\text{y})}{(1-\tan\text{y})(1-\sin\text{y}-\cos\text{y})}$
$=\lim\limits_{\text{y}\rightarrow0}\Big(\frac{-2\tan\text{y}}{(1-\tan\text{y})(1-\sin\text{y}-\cos\text{y})}\Big)$
$=-2\lim\limits_{\text{y}\rightarrow0}\frac{\tan\text{y}\times1}{\lim\limits_{\text{y}\rightarrow0}(1-\tan\text{y})\times\lim\limits_{\text{y}\rightarrow0}(1\sin\text{y}-\cos\text{y})}$
$=\frac{-2\Big(\lim\limits_{\text{y}\rightarrow0}\frac{\tan\text{y}}{\text{y}}\Big)\times\text{y}}{\Big(\lim\limits_{\text{y}\rightarrow0}(1)-\lim\limits_{\text{y}\rightarrow0}\tan\text{y}\Big)\times\Big(1-\lim\limits_{\text{y}\rightarrow0}\frac{\sin\text{y}}{\text{y}}\times\text{y}-\cos0\Big)}$
$=\frac{-2}{(1-\text{y})(1-\text{y}-1)}=\frac{-2\text{y}}{(1-\text{y})(-\text{y})}=\frac{2}{1-\text{y}}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{2}{1-\text{y}}=2$
$=2$
View full question & answer→