Limits, and Continuity - Finding whether a function is continuous or not 
$$ f(x) = \lim_{n \to \infty} \frac{\log(2 + x) - x^{2n}\sin x}{1 +
 x^{2n}} $$ 
  then, check the continuity of the function at $x = 1$.

I found this question in a text. After some thinking I figured that if $ n $ tends to $\infty$, then clearly, if we talk about $x^{2n}$, then it will approach $0$ when we try to find the left hand limit, while it will approach $\infty$ if we approach it from right hand. Is this line of reasoning correct?
From this reasoning, I am getting left hand limit as $ \log(3) $ and right hand limit as $-\sin 1$. Can someone please confirm this, and if there's any other better approach to this question, I would love to hear it.
 A: First note that the expression inside the limit is define only for $x$ such that 
$2+x>0$, that is, for $x>-2$. So let $x>-2$ be a constant. Note that if $|x|<1$, then $x^{2n}\to0$, so 
$$
 f(x) = \lim_{n \to \infty} \frac{\log(2 + x) - x^{2n}\sin x}{1 +
 x^{2n}}=\frac{\log(2 + x) -0\sin x}{1 +
 0}=\log(2 + x)
$$
If $-2<x<-1$ or $1<x$, then $x^{2n}\to\infty$, so 
$$
 f(x) = \lim_{n \to \infty} \frac{\log(2 + x) - x^{2n}\sin x}{1 +
 x^{2n}}=
\lim_{n \to \infty}\left(\frac{\log(2 + x)}{1+x^{2n}}-\frac{\sin x}{\frac{1}{x^{2n}}+1}\right)=-\sin x
$$
Finally, if $x=\pm1$ , then 
$$
 f(x) = \lim_{n \to \infty} \frac{\log(2 + x) - x^{2n}\sin x}{1 +
 x^{2n}}=
\lim_{n \to \infty} \frac{\log(2 + x) -\sin x}{1 +1}=\frac{\log(2 + x) -\sin x}{2}
$$
Therefore
$$
f(x)=
\begin{cases}
-\sin x   & -2<x<-1\ \text{or}\ 1<x\\
\log(2+x)                    &-1<x<1\\
\frac{\log(2 + x) -\sin x}{2} & x=\pm1\\
\end{cases}
$$
Note that 
$$
\lim_{x\to1^{+}}f(x)=\lim_{x\to1^{+}}(-\sin x)=-\sin1
$$
and 
$$
\lim_{x\to1^{-}}f(x)=\lim_{x\to1^{-}}\log(2+x)=\log3
$$
Since $\lim_{x\to1^{-}}f(x)\neq\lim_{x\to1^{+}}f(x)$ it follows that $f$ is not continues at $x=1$. 
A: Your approach is fine. It suffices to check for continuity whether we have 
$$
\lim_{x_n\to1}f(x_n)=f(1)
$$
for all sequences $x_n$ converging to $1$.
First we have
$$
f(1)=\lim_{n \to \infty} \frac{\log(2 + 1) - 1^{2n}\sin 1}{1 +
 1^{2n}}=\frac{\log(3)-\sin 1} 2
$$
Now we check the left limit $x_n \uparrow 1$ which is
$$
\lim_{x_n\uparrow1}f(x_n)=\lim_{x_n\uparrow1}\lim_{k \to \infty} \frac{\log(2 + x_n) - x_n^{2k}\sin x_n}{1 +
 x_n^{2k}}=\log 3
$$
and for the right limit $x_n \downarrow 1$ which is
$$
\lim_{x_n\downarrow1}f(x_n)=\lim_{x_n\downarrow1}\lim_{k \to \infty} \frac{\log(2 + x_n) - x_n^{2k}\sin x_n}{1 +
 x_n^{2k}}=-\sin1
$$
which shows that 
$$
\lim_{x_n\downarrow1}f(x_n)\neq\lim_{x_n\uparrow1}f(x_n)\neq f(1)
$$
and hence
$$
\lim_{x_n\to1}f(x_n) \text{ does not exits}
$$
and therefore $f$ is not continuous in $x_0=1$ .
