Question

If $f\left(x\right)=\underset{n\to \infty }{lim}\frac{\log (2+x)-x^{2n}\sin x}{1+x^{2n}}$, then $f\left(x\right)$ is discontinuous at

• A. $x=1$ only
• B. $x=-1$ only
• C. $x=-1,1$ only
• D. no point

Answer 1

The correct option is C $x=-1,1$ only

$f\left(x\right)={lim}_{n\to \infty }\frac{\log (2+x)-x^{2n}\sin x}{1+x^{2n}}$
If $|x|<1,x^{2n}\to 0$, as $n\to \infty$
If $|x|>1,x^{-2n}\to 0$, as $n\to \infty$
$f\left(x\right)=\{\begin{array}{c}\log (2+x);|x|<1\\ -\sin x;|x|>1\\ \frac{\log 3-\sin 1}{2};x=1\\ \frac{\sin 1}{2};x=-1\end{array}$
Clearly $f\left(1^{-}\right)\ne f\left(1^{+}\right)\ne f\left(1\right)$ and $f\left(1^{-}\right)\ne f(-1^{+})\ne f(-1)$
$\therefore f\left(x\right)$ is discontinuous at $x=\pm 1$ only.