Compute $\lim \limits_{n \to \infty} \frac{\sqrt[4]{n^4+4n}\,-\,\sqrt[3]{n^3+3n}}{\sqrt[5]{n^5+1}\,-\, \sqrt[5]{n^5+n}}\cdot \frac{1}{n^2}$ I need to compute:
$\lim \limits_{n \to \infty} \frac{\sqrt[4]{n^4+4n}\,-\,\sqrt[3]{n^3+3n}}{\sqrt[5]{n^5+1}\,-\, \sqrt[5]{n^5+n}}\cdot\frac{1}{n^2}$
I tried this
$\lim \limits_{n \to \infty} \frac{n}{n}\frac{\sqrt[4]{n^4+4n}\,-\,\sqrt[3]{n^3+3n}}{\sqrt[5]{n^5+1}\,-\, \sqrt[5]{n^5+n}}\cdot\frac{1}{n^2}=\lim \limits_{n \to \infty} \frac{\sqrt[4]{1+\frac{4}{n^3}}\,-\,\sqrt[3]{1+\frac{3}{n^2}}}{\sqrt[5]{1+\frac{1}{n^5}}\,-\, \sqrt[5]{1+\frac{1}{n^4}}}\cdot\frac{1}{n^2}$
I get stuck and I will have to probably use different method, would someone give me an advice, how  to approach to this problem?
I also thought about multiplying it by $\frac{(a+b)}{(a+b)}$, but I am not really sure how.

The answer is 5.
I expect you're supposed to use the generalized binomial theorem here: namely, $(1+x)^n = \sum_{k=0}^{\infty}\left(\frac{n^{\underline{k}}}{k!}x^k\right)$, where $n^{\underline{k}}$ means $1$ for $k=0$, or $n(n-1)(n-2)\cdots(n-k+1)$ otherwise.
$$\lim \limits_{n \to \infty} \frac{\sqrt[4]{1+\frac{4}{n^3}}\,-\,\sqrt[3]{1+\frac{3}{n^2}}}{\sqrt[5]{1+\frac{1}{n^5}}\,-\, \sqrt[5]{1+\frac{1}{n^4}}}\cdot\frac{1}{n^2} = $$
$$\lim \limits_{n \to \infty} \frac{\left(1 +\frac{1}{n^3} + \mathrm{O}(n^{-6}) \right)\,-\,\left(1+\frac{1}{n^2} + \mathrm{O}(n^{-4})\right)}{\left(1+\frac{1}{5n^5}+\mathrm{O}(n^{-10})\right)\,-\, \left({1+\frac{1}{5n^4} +\mathrm{O}(n^{-8})}\right)}\cdot\frac{1}{n^2} =$$
$$\lim \limits_{n \to \infty} \frac{5 + \mathrm{O}(n^{-3}) \,-\,5n + \mathrm{O}(n^{-1})}{1+\mathrm{O}(n^{-5})\,-\, {n +\mathrm{O}(n^{-3})}} =$$
$$\lim_{n\to\infty}\frac{5n}{n} = 5$$
