Question

Which of the following is true about
$f\left(x\right)=\{\begin{array}{cc}\frac{(x-2)}{|x-2|}\left(\frac{x^2-1}{x^2+1}\right),& x\ne 2\\ \frac{3}{5},& x=2\end{array}$

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

Answer 1

Answer: (C)

$f\left(x\right)$ has non-removable discontinuity at $x=2$.

$\underset{x\to 2^{+}}{lim}\frac{(x-2)}{|x-2|}\left(\frac{x^2-1}{x^2+1}\right)=\underset{x\to 2^{+}}{lim}\frac{(x-2)}{(x-2)}\left(\frac{x^2-1}{x^2+1}\right)$

$=\underset{x\to 2^{+}}{lim}\left(\frac{x^2-1}{x^2+1}\right)=\frac{3}{5}$

$=\underset{x\to 2^{-}}{lim}\frac{(x-2)}{|x-2|}\left(\frac{x^2-1}{x^2+1}\right)$

$=\underset{x\to 2^{-}}{lim}\frac{(x-2)}{(2-x)}\left(\frac{x^2-1}{x^2+1}\right)=-\frac{3}{5}$

Thus, L.H.L. $\ne$ R.H.L.
Hence, the function has non-removable discontinuity at $x=2$.