Question

If the function $f\left(x\right)=\left\{\begin{array}{ll}{\left(\cos x\right)}^{1/x},& x\ne 0\\ k,& x=0\end{array}\right.$ is continuous at x = 0, then the value of k is
(a) 0
(b) 1
(c) −1
(d) e.

Answer 1

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

If$f\left(x\right)$ is continuous at $x=0$, then
$\underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right)$

$$\begin{gathered} \Rightarrow \underset{x\to 0}{\lim }{\left(\cos x\right)}^{\frac{1}{x}}=k \\ \mathrm{If}\underset{x\to a}{\lim }f\left(x\right)=1\mathrm{and}\underset{x\to a}{\lim }g\left(x\right)=0,\mathrm{then} \\ \underset{x\to a}{\lim }{\left(f\left(x\right)\right)}^{g\left(x\right)}=e^{\underset{x\to a}{\lim }\left(f\left(x\right)-1\right)\times g\left(x\right)} \\ \Rightarrow e^{\underset{x\to 0}{\lim }\frac{\left(\cos x-1\right)}{x}}=k \\ \Rightarrow e^0=k\left[\because \underset{x\to 0}{\lim }\frac{\left(\cos x-1\right)}{x}=0\right] \\ \Rightarrow k=1 \end{gathered}$$