Question

If $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{1-\cos x}{x^2},& x<0\end{array}\\ \begin{array}{cc}\frac{1}{2}{e}^x,& x\ge 0\end{array}\end{array}$ then at $x=0,f$ is

• A. continuous
• B. not continuous
• C. differentiable
• D. none of these

Answer 1

Answer: (A)

continuous

To examine the continuity at $x=0,f\left(0\right)=\frac{1}{2}$

$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}\frac{1}{2}{e}^x=\frac{1}{2}$

$\underset{x\to 0^{-}}{lim}\frac{1-\cos x}{x^2}=\frac{1}{2}$

$\Rightarrow f$ is continuous.

To examine the differentiability at $x=0$

$\underset{h\to 0^{+}}{lim}\frac{f\left(h\right)-\frac{1}{2}}{h}=\underset{h\to 0^{+}}{lim}\frac{1}{2}\left(\frac{e^h-1}{h}\right)=1$

$\underset{h\to 0^{-}}{lim}\frac{f\left(h\right)-\frac{1}{2}}{h}=\underset{h\to 0^{-}}{lim}\frac{1}{h}(\frac{1-\cos h}{h^2}-\frac{1}{2})=0$

$\Rightarrow f$ is not differentiable.