Question

If the function $f\left(x\right)=\{\begin{array}{c}{\left(\cos x\right)}^{1/x}\\ k,x=0\end{array},x\ne 0$ is continuous at $x=0$ then the value of $k$ is-

• A. $1$
• B. $-1$
• C. $0$
• D. $e$

Answer 1

The correct option is A $1$

$f\left(x\right)=\{\begin{array}{c}{\left(\cos x\right)}^{1/x}\\ k,x=0\end{array},x\ne 0$

Continuous at $x=0$

$=\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}f\left(x\right)=f\left(0\right)$

$=\underset{x\to 0^{+}}{lim}f\left(x\right)$

$=\underset{x\to 0^{+}}{lim}{\left(\cos x\right)}^{1/x}$

$\underset{h\to 0}{lim}{\left(\cos h\right)}^{1/h}\Rightarrow {\left(1\right)}^{\infty }$

$=\underset{h\to 0}{lim}(cosh-1)\times \frac{1}{h}$

$h\frac{1-2{\sin }^2\frac{h}{2}-1}{h}=\frac{-2\times {\sin }^2\frac{h}{2}\times h}{\frac{h}{4}\times 4\times h}$

$=\frac{-1}{2}{\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)}^2\times h$

$=\frac{-h}{2}=0=e^0=1$

$\Rightarrow 1=k$

$\Rightarrow k=1$