Chapter 9, Problem 36RE

a.

To determine

To show: $f$ has removable discontinuity at $x=0$ .

Answer to Problem 36RE

  $f$ has a removable discontinuity at $x=0$ .

Explanation of Solution

Given information: $f\left(x\right)=x\ln x$

Formula used :

  $f$ has removable discontinuity at $x=a$ if $\underset{x\to a^{+}}{\lim }f\left(x\right)$ and $\underset{x\to a^{-}}{\lim }f\left(x\right)$ both exists but not equal to $f\left(a\right)$

Calculation:

  $\begin{array}{l}\underset{x\to 0^{-}}{\lim }f\left(x\right)\\ =\underset{x\to 0^{-}}{\lim }\frac{\ln x}{\frac{1}{x}}\left[\equiv \frac{\infty }{\infty }\right]\\ =\underset{x\to 0^{-}}{\lim }\frac{\frac{1}{x}}{\left(-\frac{1}{x^2}\right)}\text{ L' Hospital rule}\\ \text{=}\underset{x\to 0^{-}}{\lim }\left(-x\right)\\ =0\end{array}$

  $\begin{array}{l}\underset{x\to 0^{+}}{\lim }f\left(x\right)\\ =\underset{x\to 0+}{\lim }\frac{\ln x}{\frac{1}{x}}\left[\equiv \frac{\infty }{\infty }\right]\\ =\underset{x\to 0^{+}}{\lim }\frac{\frac{1}{x}}{\left(-\frac{1}{x^2}\right)}\text{ L' Hospital rule}\\ \text{=}\underset{x\to 0+}{\lim }\left(-x\right)\\ =0\end{array}$

Therefore, both right hand limit and left hand limit exist but $f\left(0\right)$ not exist.

So, $f$ has a removable discontinuity at $x=0$ .

b.

To determine

To define: $f$ at $x=0$ so that it is continuous there.

Answer to Problem 36RE

  $f\left(x\right)=0$ at $x=0$

Explanation of Solution

Given information:

  $f\left(x\right)=x\ln x$

Formula used:

L’ Hospital rule

Calculation:

  $\begin{array}{l}\underset{x\to 0}{\lim }f\left(x\right)\\ =\underset{x\to 0}{\lim }x\ln x\\ =\underset{x\to 0}{\lim }\frac{\ln c}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}\left[\equiv \frac{\infty }{\infty }\right]\\ =\underset{x\to 0}{\lim }\frac{\frac{1}{x}}{\left(-\frac{1}{x^2}\right)}\left[\text{Using L' Hospital rule}\right]\\ =0\end{array}$