Question

The function $f(x) = e^{|x|}$ is

• A. continuous everywhere but not differentiable at $x = 0$
• B. continuous and differentiable everywhere
• C. not continuous at $x = 0$
• D. None of these

Answer 1

Answer: (A)

$f \left(x\right)=e^{\left|x\right|}$
$f(x) = \begin{cases} e^{-x}, & \text{ $x < 0$ } \\[2ex] e^x, & \text{ $x \ge 0$ } \end{cases}$
$\therefore e^x$ is continuous for all $x > 0$ and $e^{-x}$ is continuous for all $x < 0$.
Now, $\displaystyle \lim_{x \to 0^-} f (x)=$ $\displaystyle \lim_{x \to 0^-}\,e^{-x}=e^0=1 $
$\displaystyle \lim_{x \to 0^+} f (x)=$$\displaystyle \lim_{x \to 0^-} e^{x}=e^0=1$
$\displaystyle \lim_{x \to 0^-} f (x)=$$\displaystyle \lim_{x \to 0^+} f (x)=$$\displaystyle \lim_{x \to 0} f (x)=f(0)$
Hence, $f(x)$ is continuous everywhere.
$Lf'(0)=\displaystyle \lim_{h \to 0^-}$ $\frac{f \left(0-h\right)-f \left(0\right)}{h}$
$=\displaystyle \lim_{h \to 0^-}$$\frac{e^{-h}-1}{h}=$$\displaystyle \lim_{h \to 0^-}$$\frac{\left(1-h+\frac{h^{2}}{2!}-\frac{h^{3}}{3!}+...\right)-1}{h}$
$=\displaystyle \lim_{h \to 0^-}$ $\left(-1+\frac{h}{2!}-\frac{h^{2}}{3!} \right)=-1$
$Rf'(0)=\displaystyle \lim_{h \to 0^+}$$\frac{f \left(0+h\right)-f \left(0\right)}{h}$$=\displaystyle \lim_{h \to 0^+}$$\frac{e^{h}-1}{h}$
$=\displaystyle \lim_{h \to 0^+}$$\frac{\left(1+h+\frac{h^{2}}{2!}+\frac{h^{3}}{3!}+\ldots\right)-1}{h}$
$=\displaystyle \lim_{h \to 0^+}$$\left(1+\frac{h}{2!}+\frac{h^{2}}{3!}+\ldots\right)=1$
$L(f'(0))\ne R(f'(0))$.
So $f(x)$ is not differentiable at $x = 0$.