Chapter 8.2, Problem 56E

To determine

To find: The function $f\left(x\right)$ has a removable discontinuity at $x=0$ and extends the definition of $f$ to $x=0$ .

Answer to Problem 56E

The function has removable discontinuity.

Explanation of Solution

Given Information:

The function is defined as,

  $f\left(x\right)={\left|x\right|}^x,x\ne 0$

Formula Used:

L’hospital rule is defined as,

If the limit is giving an undefined form $\frac{\infty }{\infty },\frac{0}{0}$ for the function $\frac{f\left(x\right)}{g\left(x\right)}$ .

  $\frac{f\left(x\right)}{g\left(x\right)}=\frac{\frac{d}{dx}\left(f\left(x\right)\right)}{\frac{d}{dx}\left(g\left(x\right)\right)}$

Calculation:

Consider the given fraction,

  $f\left(x\right)={\left|x\right|}^x,x\ne 0$

Here, the function $f\left(x\right)$ is defined at $x\ne 0$ and $\underset{x\to 0}{\lim }f\left(x\right)$ leads to the determinate form $0^0$ .

  $\begin{array}{c}\underset{x\to 0}{\lim }{\left|x\right|}^x=\underset{x\to 0}{\lim }{e}^{\ln {\left|x\right|}^x}\\ =\underset{x\to 0}{\lim }{e}^{x\ln \left|x\right|}\\ =e^{\underset{x\to 0}{\lim }x\ln \left|x\right|}\end{array}$

Further simplify.

  $\begin{array}{c}\underset{x\to 0}{\lim }x\ln \left|x\right|=\underset{x\to 0}{\lim }\frac{\ln \left|x\right|}{\frac{1}{x}}\\ =\underset{x\to 0}{\lim }\frac{\frac{1}{x}}{-\frac{1}{x^2}}\left[\text{L'hospital rule}\right]\\ =\underset{x\to 0}{\lim }\left(-x\right)\\ =0\end{array}$

Resubstitute the values.

  $\begin{array}{c}\underset{x\to 0}{\lim }{\left|x\right|}^x=e^0\\ =1\end{array}$

Therefore, the function $f\left(x\right)$ has a removable discontinuity at $x=0$ and extend the definition of function by letting $f\left(0\right)=1$ .

  $\begin{array}{c}\underset{x\to 0}{\lim }\frac{9x-3\sin 3x}{5x^3}=\underset{x\to 0}{\lim }\frac{9-9\cos 3x}{15x^2}\left[\text{L'hospital rule}\right]\\ =\underset{x\to 0}{\lim }\frac{0+27\sin 3x}{30x}\left[\text{L'hospital rule}\right]\\ =\underset{x\to 0}{\lim }\frac{81\cos 3x}{30}\left[\text{L'hospital rule}\right]\\ =\frac{27}{10}\end{array}$

Hence, the obtained value of $c$ is $\frac{27}{10}$ .