Discuss the continuity of the function

$f\left(x\right)=\underset{n\to \infty }{\lim }\frac{\log (2+x)-x^{2n}\sin x}{1+x^{2n}}\text{ at }x=1\text{. }$

We are supposed to find the continuity of the function $f\left(x\right)=\underset{n\to \infty }{\lim }\frac{\log (2+x)-x^{2n}\sin x}{1+x^{2n}}\text{ at }x=1\text{. }$

Now, we know that $\underset{n\to \infty }{\lim }{x}^{2n}=\left\{\begin{array}{l}0,\text{ if }{x}^2<1\\ \infty ,\text{ if }{x}^2>1\end{array}\right.$

Hence, to check existence of limit, we evaluate the right-hand side and left-hand side limits separately.

$\therefore$ For $x\to 1^{-}$, we have,

$$\begin{gathered} LHL=\underset{n\to \infty , x\to 1^{-}}{\lim }\frac{\log (2+x)-x^{2n}\sin x}{1+x^{2n}} \\ =\underset{x\to 1^{-}}{\lim }\frac{\log (2+x)}{1} [On neglecting the term which tends to zero in addition] \\ = \log 3 \end{gathered}$$

And now, for $x\to 1^{+}$, we have,

$$\begin{gathered} RHL=\underset{n\to \infty , x\to 1^{+}}{\lim }\frac{\log (2+x)-x^{2n}\sin x}{1+x^{2n}} \\ =\underset{n\to \infty , x\to 1^{+}}{\lim }\frac{-x^{2n}\sin x}{x^{2n}} [Neglecting finite terms in addition] \\ =\underset{x\to 1^{+}}{\lim }\sin x \\ =\sin 1 \end{gathered}$$

Thus, it can be clearly seen that $\underset{x\to 1^{+}}{\lim }f\left(x\right)\ne \underset{x\to 1^{+}}{\lim }f\left(x\right)$.

Therefore, $f\left(x\right)$ is not continuous at $x=1$ and option $1$ is correct.