<s>[INST] 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 ?
 [/INST] 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 $\frac{53}{54 e}=0.3611$
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