If the function

(b)

Given: $f(x)=\left\{\begin{array}{l}(\cos x)^{\frac{1}{x}} \\ k, x=0\end{array}, x \neq 0\right.$

If $f(x)$ is continuous at $x=0$, then

$\lim _{x \rightarrow 0} f(x)=f(0)$

$\Rightarrow \lim _{x \rightarrow 0}(\cos x)^{\frac{1}{x}}=k$

If $\lim _{x \rightarrow a} f(x)=1$ and $\lim _{x \rightarrow a} g(x)=0$, then

$\lim (f(x))^{g(x)}=e^{\lim _{x \rightarrow a}(f(x)-1) \times g(x)}$

$\Rightarrow e^{\lim _{x \rightarrow 0} \frac{(\cos x-1)}{x}}=k$

$\Rightarrow e^{0}=k \quad\left[\because \lim _{x \rightarrow 0} \frac{(\cos x-1)}{x}=0\right]$

$\Rightarrow k=1$