Question

Is the function $f\left(x\right)=\stackrel{Lt}{n\to \infty }\frac{\log (2+x)-x^{2n}\sin x}{1+x^{2n}}$ continuous at $x=1$?

• A. True
• B. False

Answer 1

The correct option is B False

$g\left(x\right)={\underset{n\to \infty }{Lt}x}^{2n}=\{\begin{array}{c}0,x<1\\ 1,x=1\\ \infty ,x>1\end{array}$
$f\left(x\right)=\underset{n\to \infty }{Lt}\frac{\log (2+x)-x^{2n}\sin x}{1+x^{2n}}=\{\begin{array}{c}\log (2+x),x<1\\ \frac{\log (2+x)-\sin x}{2},x=1\\ -\sin x,x>1\end{array}$
$f(1-)=\log 3$ and $f(1+)=-\sin 1$ and $f\left(1\right)=\frac{\log 3-\sin 1}{2}$
Since, $f(1-)\ne f\left(1\right)\ne f(1+)$
Therefore, $f$ is discontinuous at $x=1$