Question

Which of the following is true about
$f(x)=\{\begin{array}{ll}\frac{(x-2)}{|x-2|}\left(\frac{x^2-1}{x^2+1}\right)& \text{if x=2 }\\ \frac{3}{5};& \text{if x=2 }\end{array}$

• A. $f(x)$ is continuous at $x=2$
• B. $f(x)$ has removable discontinuity at $x=2$
• C. $f(x)$ has non-removable discontinuity at $x=2$
• D. Discontinuity at $x=2$ can be removed by redefining function at $x=2$

Answer 1

Answer: (C)

$\underset{x\to 2^{+}}{\lim }\frac{(x-2)}{|x-2|}\left(\frac{x^2-1}{x^2+1}\right)$
$=\underset{x\to 2^{+}}{\lim }\frac{(x-2)}{(x-2)}\left(\frac{x^2-1}{x^2+1}\right)$
$=\underset{x\to 2^{+}}{\lim }\left(\frac{x^2-1}{x^2+1}\right)=\frac{3}{5}$
$\underset{x\to 2^{-}}{\lim }\frac{(x-2)}{|x-2|}\left(\frac{x^2-1}{x^2+1}\right)$
$=\underset{x\to 2^{-}}{\lim }\frac{(x-2)}{(2-x)}\left(\frac{x^2-1}{x^2+1}\right)$
$=-\underset{x\to 2^{\ast }}{\lim }\left(\frac{x^2-1}{x^2+1}\right)=-\frac{3}{5}$
Thus, $L.H.L.\ne R.H.L.$
Hence, function has non-removable discontinuous at $x=2$