79939
\(\lim _{x \rightarrow 0}(\operatorname{cosec} x)^{1 / \log x}\) is equal to :
1 0
2 1
3 \(\frac{1}{\mathrm{e}}\)
4 None of these
Explanation:
(C) : Let, \(y=\lim _{x \rightarrow 0}(\operatorname{cosec} x)^{1 / \log x}\) Taking \(\log\) on both sides, we get \(\log y=\lim _{x \rightarrow 0} \frac{\log \operatorname{cosec} x}{\log x} \quad\left[\frac{\infty}{\infty}\right.\) form \(]\) \(\log y=\lim _{x \rightarrow 0} \frac{-\cot x}{1 / x} \quad\) (By L-Hospital rule) \(\log y=-\lim \frac{x}{\tan x}\) \(\left(\because \cot \mathrm{x}=\frac{1}{\tan \mathrm{x}}\right)\) \(\log \mathrm{y}=-1 \Rightarrow \mathrm{y}=\mathrm{e}^{-1}=\frac{1}{\mathrm{e}}\) Hence, required limit \(=\frac{1}{\mathrm{e}}\)
BITSAT-2014
Limits, Continuity and Differentiability
79940
If \(f: R \rightarrow R\) is defined by \(f(x)=\left\{\begin{array}{cr}\frac{2 \sin x-\sin 2 x}{2 x \cos x}, \text { if } x \neq 0 \\ a, \text { if } x=0\end{array}\right.\) then the value of a so that \(f\) is continuous at 0 is
1 2
2 1
3 -1
4 0
Explanation:
(D) : Given, \(f(x)=\left\{\begin{array}{ccc}\frac{2 \sin x-\sin 2 x}{2 x \cos x} \text { if } x \neq 0 \\ a \text { if } x=0\end{array}\right.\) \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{2 x \cos x} \quad\left(\frac{0}{0}\right.\) form \()\) Using L-Hospital rule, \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2 \cos x-2 \cos 2 x}{2(\cos x-x \sin x)}=\lim _{x \rightarrow 0} \frac{2-2}{2(1-0)}=0\) Since, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\) \(\therefore \mathrm{f}(0)=\lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x}) \Rightarrow \mathrm{a}=0\)
VITEEE-2009
Limits, Continuity and Differentiability
79941
The value of \(f(0)\) so that \(\frac{\left(-e^{x}+2^{x}\right)}{x}\) may be continuous at \(x=0\) is
79939
\(\lim _{x \rightarrow 0}(\operatorname{cosec} x)^{1 / \log x}\) is equal to :
1 0
2 1
3 \(\frac{1}{\mathrm{e}}\)
4 None of these
Explanation:
(C) : Let, \(y=\lim _{x \rightarrow 0}(\operatorname{cosec} x)^{1 / \log x}\) Taking \(\log\) on both sides, we get \(\log y=\lim _{x \rightarrow 0} \frac{\log \operatorname{cosec} x}{\log x} \quad\left[\frac{\infty}{\infty}\right.\) form \(]\) \(\log y=\lim _{x \rightarrow 0} \frac{-\cot x}{1 / x} \quad\) (By L-Hospital rule) \(\log y=-\lim \frac{x}{\tan x}\) \(\left(\because \cot \mathrm{x}=\frac{1}{\tan \mathrm{x}}\right)\) \(\log \mathrm{y}=-1 \Rightarrow \mathrm{y}=\mathrm{e}^{-1}=\frac{1}{\mathrm{e}}\) Hence, required limit \(=\frac{1}{\mathrm{e}}\)
BITSAT-2014
Limits, Continuity and Differentiability
79940
If \(f: R \rightarrow R\) is defined by \(f(x)=\left\{\begin{array}{cr}\frac{2 \sin x-\sin 2 x}{2 x \cos x}, \text { if } x \neq 0 \\ a, \text { if } x=0\end{array}\right.\) then the value of a so that \(f\) is continuous at 0 is
1 2
2 1
3 -1
4 0
Explanation:
(D) : Given, \(f(x)=\left\{\begin{array}{ccc}\frac{2 \sin x-\sin 2 x}{2 x \cos x} \text { if } x \neq 0 \\ a \text { if } x=0\end{array}\right.\) \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{2 x \cos x} \quad\left(\frac{0}{0}\right.\) form \()\) Using L-Hospital rule, \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2 \cos x-2 \cos 2 x}{2(\cos x-x \sin x)}=\lim _{x \rightarrow 0} \frac{2-2}{2(1-0)}=0\) Since, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\) \(\therefore \mathrm{f}(0)=\lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x}) \Rightarrow \mathrm{a}=0\)
VITEEE-2009
Limits, Continuity and Differentiability
79941
The value of \(f(0)\) so that \(\frac{\left(-e^{x}+2^{x}\right)}{x}\) may be continuous at \(x=0\) is
79939
\(\lim _{x \rightarrow 0}(\operatorname{cosec} x)^{1 / \log x}\) is equal to :
1 0
2 1
3 \(\frac{1}{\mathrm{e}}\)
4 None of these
Explanation:
(C) : Let, \(y=\lim _{x \rightarrow 0}(\operatorname{cosec} x)^{1 / \log x}\) Taking \(\log\) on both sides, we get \(\log y=\lim _{x \rightarrow 0} \frac{\log \operatorname{cosec} x}{\log x} \quad\left[\frac{\infty}{\infty}\right.\) form \(]\) \(\log y=\lim _{x \rightarrow 0} \frac{-\cot x}{1 / x} \quad\) (By L-Hospital rule) \(\log y=-\lim \frac{x}{\tan x}\) \(\left(\because \cot \mathrm{x}=\frac{1}{\tan \mathrm{x}}\right)\) \(\log \mathrm{y}=-1 \Rightarrow \mathrm{y}=\mathrm{e}^{-1}=\frac{1}{\mathrm{e}}\) Hence, required limit \(=\frac{1}{\mathrm{e}}\)
BITSAT-2014
Limits, Continuity and Differentiability
79940
If \(f: R \rightarrow R\) is defined by \(f(x)=\left\{\begin{array}{cr}\frac{2 \sin x-\sin 2 x}{2 x \cos x}, \text { if } x \neq 0 \\ a, \text { if } x=0\end{array}\right.\) then the value of a so that \(f\) is continuous at 0 is
1 2
2 1
3 -1
4 0
Explanation:
(D) : Given, \(f(x)=\left\{\begin{array}{ccc}\frac{2 \sin x-\sin 2 x}{2 x \cos x} \text { if } x \neq 0 \\ a \text { if } x=0\end{array}\right.\) \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{2 x \cos x} \quad\left(\frac{0}{0}\right.\) form \()\) Using L-Hospital rule, \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2 \cos x-2 \cos 2 x}{2(\cos x-x \sin x)}=\lim _{x \rightarrow 0} \frac{2-2}{2(1-0)}=0\) Since, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\) \(\therefore \mathrm{f}(0)=\lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x}) \Rightarrow \mathrm{a}=0\)
VITEEE-2009
Limits, Continuity and Differentiability
79941
The value of \(f(0)\) so that \(\frac{\left(-e^{x}+2^{x}\right)}{x}\) may be continuous at \(x=0\) is
79939
\(\lim _{x \rightarrow 0}(\operatorname{cosec} x)^{1 / \log x}\) is equal to :
1 0
2 1
3 \(\frac{1}{\mathrm{e}}\)
4 None of these
Explanation:
(C) : Let, \(y=\lim _{x \rightarrow 0}(\operatorname{cosec} x)^{1 / \log x}\) Taking \(\log\) on both sides, we get \(\log y=\lim _{x \rightarrow 0} \frac{\log \operatorname{cosec} x}{\log x} \quad\left[\frac{\infty}{\infty}\right.\) form \(]\) \(\log y=\lim _{x \rightarrow 0} \frac{-\cot x}{1 / x} \quad\) (By L-Hospital rule) \(\log y=-\lim \frac{x}{\tan x}\) \(\left(\because \cot \mathrm{x}=\frac{1}{\tan \mathrm{x}}\right)\) \(\log \mathrm{y}=-1 \Rightarrow \mathrm{y}=\mathrm{e}^{-1}=\frac{1}{\mathrm{e}}\) Hence, required limit \(=\frac{1}{\mathrm{e}}\)
BITSAT-2014
Limits, Continuity and Differentiability
79940
If \(f: R \rightarrow R\) is defined by \(f(x)=\left\{\begin{array}{cr}\frac{2 \sin x-\sin 2 x}{2 x \cos x}, \text { if } x \neq 0 \\ a, \text { if } x=0\end{array}\right.\) then the value of a so that \(f\) is continuous at 0 is
1 2
2 1
3 -1
4 0
Explanation:
(D) : Given, \(f(x)=\left\{\begin{array}{ccc}\frac{2 \sin x-\sin 2 x}{2 x \cos x} \text { if } x \neq 0 \\ a \text { if } x=0\end{array}\right.\) \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{2 x \cos x} \quad\left(\frac{0}{0}\right.\) form \()\) Using L-Hospital rule, \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2 \cos x-2 \cos 2 x}{2(\cos x-x \sin x)}=\lim _{x \rightarrow 0} \frac{2-2}{2(1-0)}=0\) Since, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\) \(\therefore \mathrm{f}(0)=\lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x}) \Rightarrow \mathrm{a}=0\)
VITEEE-2009
Limits, Continuity and Differentiability
79941
The value of \(f(0)\) so that \(\frac{\left(-e^{x}+2^{x}\right)}{x}\) may be continuous at \(x=0\) is
79939
\(\lim _{x \rightarrow 0}(\operatorname{cosec} x)^{1 / \log x}\) is equal to :
1 0
2 1
3 \(\frac{1}{\mathrm{e}}\)
4 None of these
Explanation:
(C) : Let, \(y=\lim _{x \rightarrow 0}(\operatorname{cosec} x)^{1 / \log x}\) Taking \(\log\) on both sides, we get \(\log y=\lim _{x \rightarrow 0} \frac{\log \operatorname{cosec} x}{\log x} \quad\left[\frac{\infty}{\infty}\right.\) form \(]\) \(\log y=\lim _{x \rightarrow 0} \frac{-\cot x}{1 / x} \quad\) (By L-Hospital rule) \(\log y=-\lim \frac{x}{\tan x}\) \(\left(\because \cot \mathrm{x}=\frac{1}{\tan \mathrm{x}}\right)\) \(\log \mathrm{y}=-1 \Rightarrow \mathrm{y}=\mathrm{e}^{-1}=\frac{1}{\mathrm{e}}\) Hence, required limit \(=\frac{1}{\mathrm{e}}\)
BITSAT-2014
Limits, Continuity and Differentiability
79940
If \(f: R \rightarrow R\) is defined by \(f(x)=\left\{\begin{array}{cr}\frac{2 \sin x-\sin 2 x}{2 x \cos x}, \text { if } x \neq 0 \\ a, \text { if } x=0\end{array}\right.\) then the value of a so that \(f\) is continuous at 0 is
1 2
2 1
3 -1
4 0
Explanation:
(D) : Given, \(f(x)=\left\{\begin{array}{ccc}\frac{2 \sin x-\sin 2 x}{2 x \cos x} \text { if } x \neq 0 \\ a \text { if } x=0\end{array}\right.\) \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{2 x \cos x} \quad\left(\frac{0}{0}\right.\) form \()\) Using L-Hospital rule, \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2 \cos x-2 \cos 2 x}{2(\cos x-x \sin x)}=\lim _{x \rightarrow 0} \frac{2-2}{2(1-0)}=0\) Since, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\) \(\therefore \mathrm{f}(0)=\lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x}) \Rightarrow \mathrm{a}=0\)
VITEEE-2009
Limits, Continuity and Differentiability
79941
The value of \(f(0)\) so that \(\frac{\left(-e^{x}+2^{x}\right)}{x}\) may be continuous at \(x=0\) is
