A Limit problem : No-existence I find difficulty proving the no existence of this limit
I show my process
$$ \lim_{x\to 0} \biggl(1 + x e^{-  \frac{1}{x^2}}+\sin \frac{1}{x^4}\biggr)^{e^{\frac{1}{x^2}}}$$
We begin with rewriting the limit as follows:
$$ \lim_{x\to 0} \biggl(1 + x e^{-  \frac{1}{x^2}}+\sin \frac{1}{x^4}\biggr)^{e^{\frac{1}{x^2}}}=\lim_{x\to 0} \biggl( 1 + x  \frac{1}{e^{\frac{1}{x^2}}} +\sin \frac{1}{x^4}\biggr)^{e^{\frac{1}{x^2}}}$$
and analyze the various addends and the exponent of the limit:
$$\begin{align*}
&x  \frac{1}{e^{\frac{1}{x^2}}}\to 0\\
&\sin \frac{1}{x^4}\to \not \exists\\
&e^{\frac{1}{x^2}}\to+\infty.\\
\end{align*}$$
The problem here lies in the fact that we have an addendum that there is no limit, let's consider:
$$\sin{a_n}\quad\text{e}\quad\sin{b_n}\quad\text{con }\quad n\to+\infty$$
where the two sequences are:
$$a_n=\frac{\pi}{2}+2n\pi\quad\text{e}\quad b_n=2n\pi$$
Then, the function values ​​calculated in the sequence $ a_n $, with $ k $ positive integer, tends to $ 1 $, calculated values ​​of the sequence $b_n$ tends to $0$,:
$$\lim_ {n \to\infty}\sin{a_n}=1 \quad\text{and}\quad \lim_{n\to \infty}\sin {b_n} = 0$$
and therefore, as we know, the limit of $\sin x$ ($x \to \infty$) not exists.
Now, to prove that the given limit does not exist,i continued in this way 
$t= \frac{1}{x^2},$ (if $x\to0 \rightarrow t\to+\infty$) :
$$\lim_{x\to 0} \biggl( 1 + x  \frac{1}{e^{\frac{1}{x^2}}} +\sin \frac{1}{x^4}\biggr)^{e^{\frac{1}{x^2}}}=\lim_{t\to +\infty} \biggl( 1 + \frac{  \sqrt{t}}{t}\cdot \frac{1}{e^t} +\sin{t^2}\biggr)^{e^t}$$
$\frac{  \sqrt{t}}{t}\cdot \frac{1}{e^t}\to 0$
i consider
$$\begin{align*}
&\lim_{t\to +\infty} \biggl(1 + \sin{({a_n})^2}\biggr)^{e^t}=\biggl(1+1\biggr)^{e^t}=+\infty\\
&\lim_{t\to +\infty} \biggl( 1 + \sin{({b_n})^2}\biggr)^{e^t}=e^{e^t\ln\biggl( 1 + \sin{({b_n})^2}\biggr)}=e^{+\infty\ln( 1 + 0)}=???
\end{align*}$$
 A: Hint: Better consider the subsequence where $\sin(x)=-1$ instead of the one where $\sin(x)=0$, it is easy to see that the limit is $0$ there and then you found a subsequence that converges to $0$ and one that converges to $+\infty$ (the one where $\sin(x)=1$). Hence the limit is not existent.
A: Given an integer $k$, let $\frac{1}{x^4}=2\pi k + \frac{\pi}{2}$, or $x = (2\pi k + \frac{\pi}{2})^{-\frac{1}{4}}$.  Then $\sin {\frac{1}{x^4}} = 1$. Let $y=e^{\frac{1}{x^2}}$.  Then your expression is:
$$(2+\frac{x}{y})^y$$.
Now, if $|x|<1$ then $y>e$, so $\frac{x}{y}> \frac{-1}{2}$.  So this expression is at least as big as $(\frac{3}2)^y$.
Picking a large enough $k$, we can make $x$ arbitrarily small, so $\frac{1}{x^2}$ arbitrarily large, so $y=e^{\frac{1}{x^2}}$ arbitrarily large.  So, we see that we can make your expression bigger than $(\frac{3}2)^Y$ for arbitrarily large $Y$, and hence it has no limit.
A: You can show that $$\lim_{t\to +\infty} \biggl( 1 + \frac{  \sqrt{t}}{t}\cdot \frac{1}{e^t} +\sin{t^2}\biggr)^{e^t}$$ does not exist by observing that for arbitrarily large $t_0$, there is a $t>t_0$ such that $\sin t^2 = 1$ so the value at $t'$ is at least $2^{e^{t}}$, hence we have divergence.
