Find : $\lim\limits_{n\to +\infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$ Find : 
$$\lim\limits_{n\to +\infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$$
My attempt : i don't know is correct or no!
I use this rule : 
$$\lim\limits_{n\to +\infty}(f(x))^{g(x)}=1^{\infty}$$
Then : 
$$\lim\limits_{n\to +\infty}(f(x))^{g(x)}=\lim\limits_{n\to +\infty}e^{g(x)(f(x)-1)}$$
So :
$$\lim\limits_{n\to +\infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$$
$$=\lim\limits_{n\to +\infty}\frac{e^{\frac{n^{4}}{n^{3}}}}{e^{\frac{(n+1)^{4}}{(n+1)^{3}}}}$$
$$=\lim\limits_{n\to +\infty}\frac{e^{n}}{e^{n+1}}$$
$$=\frac{1}{e}$$
Is my approach wrong ?
is this called partial limit calculation ?
 A: HINT:
Use the inequality
$$\left(1+\frac{1}{n}\right)^n < e < \left(1+\frac{1}{n}\right)^{n+1}$$
so
$$e^{\frac{1}{n+1}}< \left(1+\frac{1}{n}\right)< e^{\frac{1}{n}}$$
We get lower and upper bounds:
$$e^{\frac{n^4}{n^3+1}}< \left(1+\frac{1}{n^3}\right)^{n^4}< e^n $$
$$e^{\frac{(n+1)^4}{(n+1)^3 +1}}< \left(1 + \frac{1}{(n+1)^3}\right)^{(n+1)^4} < e^{n+1}$$
Conclude:
$$e^{\frac{n^4}{n^3+1}}/e^{n+1}< \ldots < e^n/e^{\frac{(n+1)^4}{(n+1)^3 +1}}$$
so the limit is $\frac{1}{e}$.
A: Your approach leads to the right answer, but there may be some questionable moves there. Me, I would apply $\ln$ to the expression and use
$$\tag 1 \ln (1+u) = u +O(u^2)$$
as $u\to 0.$ Thus in our problem we would get
$$n^4\ln(1+1/n^3) - (n+1)^4 \ln (1 +1/(n+1)^3).$$
Apply $(1)$ to get the answer.
A: Let's note that exponent $(n+1)^4$ in the denominator can be written as $$n^4+4n^3+\dots$$ and hence the denominator can be written as a product of a finite number of factors the first of which is $(1+(n+1)^{-3})^{n^4}$, the second of which is $(1+(n+1)^{-3})^{4n^3}$  and the remaining factors tend to $1$ (justify this yourself). The second factor tends to $e^{4}$. Thus the desired limit is equal to the limit of expression $$e^{-4}\left(\dfrac{1+\dfrac{1}{n^3}}{1+\dfrac{1}{(n+1)^3}}\right)^{n^4}$$ which can be further rewritten as $$ e^{-4}\left(1+\frac{3n^2+3n+1}{n^3(1+(n+1)^3)}\right)^{n^4}$$ and then you can apply your formula to get the limit as $$e^{-4}\exp\left(\lim_{n\to\infty} n^4\cdot\frac{3n^2+3n+1}{n^3((n+1)^3+1)}\right)=\frac{1}{e}$$

Expanding on my comment to your question, let's note that if your rule gave finite results for both numerator and denominator (including non-zero denominator) it would have been fine. But here it leads to the conclusion that both numerator and denominator tend to $\infty$. The rule says that if $f(x) \to 1$ and $g(x) \to\infty $ and further $\exp(g(x) (f(x) - 1))$ tends to a limit or to $\infty $ then so does $\{f(x) \} ^{g(x)} $. It does not mean that under these circumstances you can replace the expression $f^g$ by $\exp(g(x)(f(x)-1))$.

Note: The previous version of this answer used unnecessary manipulation which was tedious. Simplicity eludes us in surprising ways.
A: Assuming that you would like to get more than the limit itself, starting with
$$a_n=\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$$ take logarithms
$$\log(a_n)=n^4\log\left(1+\frac{1}{n^3}\right)-(n+1)^4\log\left(1+\frac{1}{(n+1)^3}\right)$$ Use twice the expansion
$$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+O\left(x^5\right)$$ and expand to get
$$\log(a_n)=-1-\frac{1}{n^3}+\frac{3}{2n^4}+O\left(\frac{1}{n^5}\right)$$ and continue with Taylor series
$$a_n=e^{\log(a_n)}=e\left(1-\frac{1}{n^3}+\frac{3}{2n^4}\right)+O\left(\frac{1}{n^5}\right)$$ which shows the limit and how it is approached. Moreover, this gives a shortcut estimation of $a_n$.
Suppose that, using your pocket calculator, you compute $a_3$. You should get
$a_3=0.3594$ while the above truncated expansion gives $\frac{53}{54 e}=0.3611$
A: Aonther method by Lagrange's MVT. Let 
$f(x)=x^4\log\left(1+\frac{1}{x^3}\right)$,then
$$\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}
=e^{f(n)-f(n+1)}.$$
By Lagrange's MVT, there exists $\xi_n\in(n,n+1)$ such that
$$f(n+1)-f(n)=f'(\xi_n).$$
An elementary computation gives that
$$f'(x)=4x^3\log\left(1+\frac{1}{x^3}\right)-\frac{3x^3}{x^3+1},$$
and $$\lim_{x\to\infty}f'(x)=4-3=1.$$
So
$$\lim_{n\to \infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}=\lim_{n\to \infty}e^{-f'(\xi_n)}=\lim_{x\to \infty}e^{-f'(x)}=\frac{1}{e}.$$
A: The problem is that you haven't shown how exactly you're using the lemma / theorem you referenced (for limits of the form $1^{\infty}$). The given limit is not $1^{\infty}$. It's actually $\infty/\infty$. So one explanation is that you're assuming
$$\lim\limits_{n\to +\infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}
=\frac{\lim_{n\to\infty}\left(1+\frac{1}{n^3}\right)^{n^4}}{\lim_{n\to\infty}\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}
$$
Now each of the limits (in the numerator and denominator) is $1^{\infty}$ and you could possibly apply the lemma. Except the above manipulation is wrong since both limits are $\infty$. What we could do instead is claim that
$$\left(1+\frac{1}{n^3}\right)^{n^4}\sim e^n\,\,(n\to\infty)\\
\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}\sim e^{n+1}\,\,(n\to\infty) $$
Such a claim requires justification of course. For example, by considering the series expansions of the expressions.
A: Hint:
$$\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}=\frac{\left(\left(1+\frac{1}{n^3}\right)^{n^3}\right)^n}{\left(\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^3}\right)^{(n+1)}}$$
$$=\frac{\left(\left(1+\frac{1}{n^3}\right)^{n^3}\right)^n}{\left(\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^3}\right)^{n}\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^3}}$$
