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

Question

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 $1$.

Answer 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.

Answer 2

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, $\lambda=\displaystyle\lim_{n\to \infty}\frac{n}{\sqrt[3]{n^3-n^2}}=\ldots=1, \text{ so } L=e^1=e.$

Answer 3

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.

Answer 4

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 $e^1=e$.