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If the function $f(x)=\left\{\begin{array}{1 1}(\cos x)^{1/x}&x\neq 0\\k&x=0\end{array}\right.$ is continuous at $x=0$ then value of $K$ is

$(a)\;1\qquad(b)\;-1\qquad(c)\;0\qquad(d)\;e$

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For continuity limit=value=K
$\therefore \lim\limits_{x\to 0}(\cos x)^{1/x}=K$
$\Rightarrow \log\lim\limits_{x\to 0}(\cos x)^{1/x}=\log K$
$\Rightarrow \lim\limits_{x\to 0}\log(\cos x)^{1/x}=\log K$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{\log\cos x}{x}$$=\log K$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{\Large\frac{1}{\cos x}(\large-\sin x)}{1}$$=\log K$ [Using L Hospital's Rule]
$\Rightarrow 0=\log K$
$\Rightarrow K=e^0$
$\Rightarrow 1$
Hence (a) is the correct answer.
answered Jan 6, 2014 by sreemathi.v
 

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