Prove that the limit $\lim \limits_{n \to \infty}\left(1 + \frac{1}{\sqrt[3]{n^3 - n^2}}\right)^n$ doesn't exist or find it Given $\lim \limits_{n \to \infty}\left(1 + \frac{1}{\sqrt[3]{n^3 - n^2}}\right)^n$.
My attempt: $\lim \limits_{n \to \infty}\left(1 + \frac{1}{\sqrt[3]{n^3 - n^2}}\right)^n = \lim \limits_{n \to \infty}\left(1 + \frac{1}{n\sqrt[3]{1 - 1/n}}\right)^n$. So as the  $n \to \infty \Rightarrow \sqrt[3]{1 - 1/n} \to 1$. 
$\lim \limits_{n \to \infty}\left(1 + \frac{1}{n\sqrt[3]{1 - 1/n}}\right)^n = \lim \limits_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e$.
However, I'm not sure that my proof is strict enough, especially the moment, where I go from $\sqrt[3]{1 - 1/n}$ to $1$.


$\lim \limits_{n \to \infty}\left(1 + \frac{1}{n\sqrt[3]{1 - 1/n}}\right)^n = \lim \limits_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e$.
However, I'm not sure that my proof is strict enough, especially the moment, where I go from $\sqrt[3]{1 - 1/n}$ to $1$.

Indeed, this step is not correct, and you need to justify why this is fine (under other circumstances, very similar-looking, it may not give the right result).
As almost always there is a doubt or some non-trivial step to be made, in this situation I would suggest to rewrite the quantity in the exponential form (below, very detailed derivation):
$$
\left(1 + \frac{1}{n\sqrt[3]{1 - 1/n}}\right)^n
= \exp\left(n \ln \left(1 + \frac{1}{n\sqrt[3]{1 - 1/n}}\right) \right)
$$
Now,
$$\begin{align}
n \ln \left(1 + \frac{1}{n\sqrt[3]{1 - 1/n}}\right)
&= \frac{1}{\sqrt[3]{1 - 1/n}}\cdot  n\sqrt[3]{1 - 1/n} \ln \left(1 + \frac{1}{n\sqrt[3]{1 - 1/n}}\right)\\
&= \frac{1}{\sqrt[3]{1 - 1/n}}\cdot  \frac{\ln \left(1 + a_n\right)}{a_n}
\end{align}$$
setting $a_n\stackrel{\rm def}{=}\frac{1}{n\sqrt[3]{1 - 1/n}}\xrightarrow[n\to\infty]{}0$. Recalling that $\frac{\ln(1+u)}{u}\xrightarrow[u\to0]{}1$, we get that
$$
\frac{1}{\sqrt[3]{1 - 1/n}}\cdot  \frac{\ln \left(1 + a_n\right)}{a_n}
\xrightarrow[n\to\infty]{}1\cdot 1=1
$$
and therefore
$$
\exp\left(\frac{1}{\sqrt[3]{1 - 1/n}}\cdot  \frac{\ln \left(1 + a_n\right)}{a_n}\right)
\xrightarrow[n\to\infty]{} e^1 = e
$$
by continuity of the exponential.
