Question

For the function $f\left(x\right)=\frac{1}{\ln \left|x\right|}$ which of the following statements is/are true

• A. $f\left(x\right)$ has removable discontinuity at $x=0$
• B. $f\left(x\right)$ has non-removable discontinuity at $x=1$
• C. $f\left(x\right)$ has removable discontinuity at $x=-1$
• D. $f\left(x\right)$ has non-removable discontinuity at $x=0$

Answer 1

Answer: (A), (B)

$f\left(x\right)$ has non-removable discontinuity at $x=1$

$f\left(x\right)=\{\begin{array}{cc}\frac{1}{\ln x}& \text{if}x>0,x\ne 1\\ \frac{1}{\ln (-x)}& \text{if}x<0,x\ne -1\end{array}$ function is obviously discontinuous at $x=0,1,-1$ as it is not defined.
At $x=0$
$\begin{array}{cc}\underset{x\to 0^{+}}{lim}f\left(x\right)=0& \\ \underset{x\to 0^{-}}{lim}f\left(x\right)=0& \end{array}\}$
Limits exists at $x=0$. Hence $f\left(x\right)$ has removable discontinuity at $x=0$. (Missing point discontinuity)
At $x=1$
$\begin{array}{cc}\underset{x\to 1^{+}}{lim}f\left(x\right)=\infty & \\ \underset{x\to 1^{-}}{lim}f\left(x\right)=-\infty & \end{array}\}$
Hence $f\left(x\right)$ has non-removable discontinuity (infinite type) at $x=1$
At $x=-1$
$\begin{array}{cc}\underset{x\to -1^{+}}{lim}f\left(x\right)=-\infty & \\ \underset{x\to -1^{-}}{lim}f\left(x\right)=\infty & \end{array}\}$
Hence $f\left(x\right)$ has non-removable discontinuity (infinite type) at $x=-1$