Chapter 4.5, Problem 76E

a.

To determine

To show that f is continuous at $0$ .

Explanation of Solution

Given:

The given function is

  $f(x)=\left\{\begin{array}{l}{\left|x\right|}^x\text{if }x\ne 0\\ 1\text{if }x=0\end{array}\right\}$

Calculation:

For $x<0$ the function can be written as

  $f(x)={(-x)}^x$

This implies that −

  $\begin{array}{l}$ y=\underset{x\to 0^{-}}{\lim }{(-x)}^x$$ \ln y=\underset{x\to 0^{-}}{\lim }\ln {(-x)}^x\{\because L\text{ Hopital's Rule}}$\ln y=\underset{x\to 0^{-}}{\lim }x\ln (-x)\\ \ln y=\underset{x\to 0^{-}}{\lim }\frac{\ln (-x)}{\frac{1}{x}}\\ \ln y=0\\ y=e^0=1\end{array}$

For $x>0$ the function can be written as

  $f(x)={(x)}^x$

This implies that −

  $\begin{array}{l}$ y=\underset{x\to 0^{+}}{\lim }{(x)}^x$$ \ln y=\underset{x\to 0^{+}}{\lim }\ln {(x)}^x\{\because L\text{ Hopital's Rule}}$\ln y=\underset{x\to 0^{+}}{\lim }x\ln (x)\\ \ln y=\underset{x\to 0^{+}}{\lim }\frac{\ln (x)}{\frac{1}{x}}\\ \ln y=0\\ y=e^0=1\end{array}$

Since the function approaches $1$ from both sides, therefore it is continuous at $x=0$

b.

To determine

To investigate graphically whether $f$ is differentiable at $x=0$

Explanation of Solution

Given:

The given function is

  $f(x)=\left\{\begin{array}{l}{\left|x\right|}^x\text{if }x\ne 0\\ 1\text{if }x=0\end{array}\right\}$

Calculation:

  

  

c.

To determine

To show that $f$ is not differentiable at $x=0$ using the graphs in part (b).

Explanation of Solution

Given:

The given function is

  $f(x)=\left\{\begin{array}{l}{\left|x\right|}^x\text{if }x\ne 0\\ 1\text{if }x=0\end{array}\right\}$

Calculation:

  $\begin{array}{l}\ln f(x)=x\ln \left|x\right|\\ f\text{'}(x)=f(x)\cdot \frac{f\text{'}(x)}{f(x)}=f(x)(1+\ln |x|)\text{ and }x\ne 0\\ \Rightarrow x\to 0,f\text{'}(x)\to -\infty ,\end{array}$

So, there is a vertical tangent at $(0,1)$ and the function is not differentiable at this point.