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 $1$.
 A: 
$\lim \limits_{n \to \infty}\left(1 + \frac{1}{n\sqrt[3]{1 - 1/n}}\right)^n = \lim \limits_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e$.
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.
A: The end result is correct, but your last step lacks rigour. Since our limit is of the form "$1^\infty$", the usual trick will do:
$$\lim \limits_{n \to \infty} \left( 1 + \frac 1 {\sqrt[3] {n^3 - n^2}} \right) ^n = \lim \limits_{n \to \infty} \left[ \left( 1 + \frac 1 {\sqrt[3] {n^3 - n^2}} \right) ^{\sqrt[3] {n^3 - n^2}} \right] ^{\frac n {\sqrt[3] {n^3 - n^2}}} = \\
\left[ \lim \limits_{n \to \infty} \left( 1 + \frac 1 {\sqrt[3] {n^3 - n^2}} \right) ^{\sqrt[3] {n^3 - n^2}} \right] ^{\lim \limits_{n \to \infty} \frac n {\sqrt[3] {n^3 - n^2}}} = \Bbb e ^{\lim \limits_{n \to \infty} \frac n {\sqrt[3] {n^3 - n^2}}} = \Bbb e ^1 = \Bbb e.$$
"The usual trick" that I have mentioned is the following: whenever you have to study a limit of the type $\left( 1 + \frac 1 {x_n} \right) ^{y_n}$ with $x_n, y_n \to \infty$, rewrite it as
$$\left[ \left( 1 + \frac 1 {x_n} \right) ^{x_n} \right] ^{\frac {y_n} {x_n}}$$
and use the fact that $\left( 1 + \frac 1 {x_n} \right) ^{x_n} \to \Bbb e$. The whole difficulty of the exrcise becomes, then, the computation of $\lim \frac {y_n} {x_n}$ which in many cases is considerably simpler than any other approach that you might consider.
A: The last equality should be explained. At any case we can use that if $a_n\to 1$ and $b_n\to \infty$ then, 
$$L=\displaystyle\lim_{n\to \infty}a_n^{b_n}=e^{\lambda} \text{ where } \lambda=\displaystyle\lim_{n\to \infty}(a_n-1){b_n}.$$ In our case, $\lambda=\displaystyle\lim_{n\to \infty}\frac{n}{\sqrt[3]{n^3-n^2}}=\ldots=1, \text{ so } L=e^1=e.$ 
A: Indeed, the step you're doubtful upon is a quite a dangerous jump.
You can use functions rather than sequences, if possible. And, with exponentials with variable basis, pass to logarithms. If you find the limit of the logarithm to be $l$, then your limit is $e^l$ (or $0$ if $l=-\infty$, or $\infty$ if $l=\infty$). If the function has limit at $\infty$, the sequence shares it.
Thus you need
$$
\lim_{x\to\infty}x\log\left(1 + \frac{1}{\sqrt[3]{x^3 - x^2}}\right)
$$
Now, this limit (provided it exists) is the same as
$$
\lim_{t\to0^+}\frac{1}{t}\log\left(1+\frac{t}{\sqrt[3]{1-t}}\right)=
\lim_{t\to0^+}\frac{\dfrac{t}{\sqrt[3]{1-t}}+o(t)}{t}=1
$$
Without Taylor, set $\sqrt[3]{1-t}=u$, so the limit becomes
$$
\lim_{u\to1^-}\frac{\log(1+u-u^3)-\log u}{1-u^3}=
\lim_{u\to1^-}\frac{\log u-\log(1+u-u^3)}{u-1}\frac{1}{u^2+u+1}
$$
The second factor has limit $1/3$; the first factor provides the derivative at $1$ of the function $f(u)=\log u-\log(1+u-u^3)$; since
$$
f'(u)=\frac{1}{u}-\frac{1-3u^2}{1+u-u^3}
$$
we get
$$
f'(1)=1-\frac{1-3}{1+1-1}=3
$$
and so the limit is $1$.
Finally, the sought limit is $e^1=e$.
