Question

If the function \[f\left( x \right) = \begin{cases}\left( \cos x \right)^{1/x} , & x \neq 0 \\ k , & x = 0\end{cases}\] is continuous at x = 0, then the value of k is

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Answer 1

Given: 

\[f\left( x \right) = \binom{ \left( \ cosx \right)^\frac{1}{x} }{k, x = 0}, x \neq 0\]

If  \[f\left( x \right)\]  is continuous at  \[x = 0\], then

 

\[\lim_{x \to 0} f\left( x \right) = f\left( 0 \right)\]

\[\Rightarrow \lim_{x \to 0} \left( \cos x \right)^\frac{1}{x} = k\]
\[ \text{ If } \lim_{x \to a} f\left( x \right) = 1 \text{ and } \lim_{x \to a} g\left( x \right) = 0, \text{ then } \]
\[ \lim_{x \to a} \left( f\left( x \right) \right)^{g\left( x \right)} = e^\lim_{x \to a} \left( f\left( x \right) - 1 \right) \times g\left( x \right) \]
\[ \Rightarrow e^\lim_{x \to 0} \frac{\left( \cos x - 1 \right)}{x} = k\]
\[ \Rightarrow e^0 = k \left[ \because \lim_{x \to 0} \frac{\left( \cos x - 1 \right)}{x} = 0 \right]\]
\[ \Rightarrow k = 1\]