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 $\begin{align*} &\lim_{t\to +\infty} \biggl(1 + \sin{({a_n})^2}\biggr)^{e^t}=\biggl(1+1\biggr)^{e^t}=+\infty\\ &\lim_{t\to +\infty} \biggl( 1 + \sin{({b_n})^2}\biggr)^{e^t}=e^{e^t\ln\biggl( 1 + \sin{({b_n})^2}\biggr)}=e^{+\infty\ln( 1 + 0)}=??? \end{align*}$