Question

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

• 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)

$\displaystyle\lim _{x \rightarrow 2^{+}} \frac{(x-2)}{|x-2|}\left(\frac{x^{2}-1}{x^{2}+1}\right)$
$=\displaystyle\lim _{x \rightarrow 2^{+}} \frac{(x-2)}{(x-2)}\left(\frac{x^{2}-1}{x^{2}+1}\right)$
$=\displaystyle\lim _{x \rightarrow 2^{+}}\left(\frac{x^{2}-1}{x^{2}+1}\right)=\frac{3}{5}$
$\displaystyle\lim _{x \rightarrow 2^{-}} \frac{(x-2)}{|x-2|}\left(\frac{x^{2}-1}{x^{2}+1}\right)$
$=\displaystyle\lim _{x \rightarrow 2^{-}} \frac{(x-2)}{(2-x)}\left(\frac{x^{2}-1}{x^{2}+1}\right)$
$=-\displaystyle\lim _{x \rightarrow 2^{*}}\left(\frac{x^{2}-1}{x^{2}+1}\right)=-\frac{3}{5}$
Thus, $L.H.L. \neq R.H.L. $
Hence, function has non-removable discontinuous at $x=2$