Checking if limit exists Check if limits exists $$\large\lim_{x \to \infty} \frac{2+2x+\sin2x}{(2x+\sin2x)e^{sinx}}$$
My approach to this problem was $$\large\lim_{x \to \infty} \frac{\frac{2}{x}+2+\frac{\sin2x}{x}}{(2+\frac{\sin2x}{x})e^{sinx}}$$
on further simplifying $$\large\lim_{x \to \infty} \frac{0 + 2 + 0}{(2+0)(\text{value} \; \text{between}\; \frac{1}{e} \;to \; e) }\;\;\;\; (\text{since as } \; x\to \infty \; \ sinx \; \in ( -1,1))$$
which is equal to $$\large\lim_{x \to \infty} \frac{1}{\text{value} \; \text{between}\; \frac{1}{e} \;to \; e}$$
which shows limit exist.  But  answer key says limit does not exist.
Did I do some mistake or answer key is wrong?

and welcome, so according to my results the limit when $x\to +\infty$ doesn't exist, for the following reason :
Let our function be :
$$\varphi (x) =\frac{2x+2+\sin (2x)}{(2x+\sin (2x))e^{\sin(2x)}}$$
Well first let's simplify the function :
\begin{align}
\frac{2x+2+\sin (2x)}{(2x+\sin (2x))e^{\sin(2x)}} &=\left(\frac{2x+\sin(2x)}{2x+\sin(2x)} +\frac{2}{2x+\sin(2x)}\right) \times \frac{1}{e^{\sin(x)}}\\
& = \frac{1}{e^{\sin(x)}}+ \frac{2}{(2x+\sin(2x))e^{\sin (x)} }
\end{align}
Now let's see if we can determine two functions $\psi$ and $\eta$ such that $\psi\leq \varphi\leq \eta$ :
We have :
$$-1\leq\sin(2x)\leq 1\Longleftrightarrow 2x-1\leq 2x+\sin(2x)\leq 2x+1$$
Hence :
$$1+\frac{2}{2x+1}\leq 1+ \frac{2}{2x+\sin(2x)}\leq 1+\frac{2}{2x-1}$$
And we have :
$$e^{-1}\leq e^{-\sin(x)}\leq e$$
Therefore :
$$e^{-1}+\frac{2e^{-1}}{2x+1}\leq e^{-\sin(x)}\left(1+ \frac{2}{2x+\sin(2x)}\right)\leq e+\frac{2e}{2x-1}$$
And we obtained our two functions :
$$\lim_{x\to +\infty} \psi (x) =e^{-1} \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \lim_{x\to +\infty} \eta(x) =e$$
Hence we can't apply the squeeze theorem so there's no limit when $x\to \pm \infty$, the function is just oscillating between $e^{-1}$ and $e$, you can use Desmos to visualize this.
