Chapter 9.2, Problem 56E

To determine

$f\left(x\right)$ has removable discontinuity at $x=0$ and extended function of $f$ such that $f$ is continuous there.

Answer to Problem 56E

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

Explanation of Solution

Given information : $\begin{array}{l}f\left(x\right)={\left|x\right|}^x\text{ ,x}\ne \text{0}\\ \end{array}$

Formula used : A function is said to have removable discontinuity if both the left hand limit and right hand limit exists but not equal to limiting value at a point.

Calculation :

  $\begin{array}{l}y=f\left(x\right)\text{ say }y\Rightarrow {\left|x\right|}^x\\ \Rightarrow \log y=x\log \left|x\right|\\ \text{Now, }LHL\Rightarrow \underset{x\to 0^{-}}{\lim }\left(\log y\right)\text{=}\underset{x\to 0^{-}}{\lim }x\ln \left|x\right|\\ \text{=}\underset{x\to 0^{-}}{\lim }\frac{x\ln \left(-x\right)}{1}\\ \text{=}\underset{x\to 0^{-}}{\lim }\frac{\ln \left(-x\right)}{\frac{1}{x}}\left[\equiv \frac{\infty }{\infty }\right]\\ \text{=}\underset{x\to 0^{-}}{\lim }\frac{\frac{1}{x}}{\frac{1}{x^2}}\\ =\underset{x\to 0^{-}}{\lim }\left(x\right)=0\\ \Rightarrow \log \left(\underset{x\to 0^{-}}{\lim }\right)=0\Rightarrow \underset{x\to 0^{-}}{\lim y}=1\Rightarrow \underset{x\to 0^{-}}{\lim }{\left|x\right|}^x\text{=1 }\end{array}$

  $\begin{array}{l}\text{Now, RHL}\Rightarrow \underset{x\to 0^{+}}{\lim }\left(\log y\right)\text{=}\underset{x\to 0^{+}}{\lim }x\ln \left|x\right|\\ \text{=}\underset{x\to 0^{+}}{\lim }\frac{\ln x}{\frac{1}{x}}\left[\equiv \frac{\infty }{\infty }\right]\\ \text{=}\underset{x\to 0^{+}}{\lim }\frac{\frac{1}{x}}{-\frac{1}{x^2}}\\ \underset{x\to 0^{+}}{\lim }\left(-x\right)=0\Rightarrow \underset{x\to 0^{+}}{\lim }y=e^0=1\\ \Rightarrow \underset{x\to 0^{+}}{\lim }{\left|x\right|}^x=1\end{array}$

But $f\left(0\right)$ is undefined

Therefore, at $x=0$ there is a point of discontinuity which is removable.

Now,

  $\begin{array}{l}\underset{x\to 0}{\lim }\log y=\underset{x\to 0}{\lim }x\ln \left|x\right|=\underset{x\to 0}{\lim }\frac{\ln \left|xx\right|}{\frac{1}{x}}=\underset{x\to 0}{\lim }\frac{\frac{1}{x}}{-\frac{1}{x^2}}=\underset{x\to 0}{\lim }\left(-x\right)=0\\ \Rightarrow \log \left(\underset{x\to 0}{\lim y}\right)=0\\ \Rightarrow \underset{x\to 0}{\lim }y=1\\ \Rightarrow \underset{x\to 0}{\lim }{\left|x\right|}^x=1\end{array}$

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