<s>[INST] 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.
 [/INST] I know you are all "begeistert" with taylor series, but here is another method.
Note that $$\sqrt[3]{n^3+3n}-n=\frac{3n}{(\sqrt[3]{n^3+3n})^2+(\sqrt[3]{n^3+3n})n+n^2}$$ so it follows easily that $$n(\sqrt[3]{n^3+3n}-n)\to \frac{3}{3}=1$$
in a similar fashion one sees that 
$$n^2(\sqrt[4]{n^4+4n}-n)\to 1$$
$$n^4(\sqrt[5]{n^5+1}-n)\to \frac{1}{5}$$
$$n^3(\sqrt[5]{n^5+n}-n)\to \frac{1}{5}$$
Now write the original expression as
$$\frac{\frac{n^2(\sqrt[4]{n^4+4n}-n)}{n}-n(\sqrt[3]{n^3+3n}-n)}{\frac{n^4(\sqrt[5]{n^5+1}-n)}{n}-n^3(\sqrt[5]{n^5+n}-n)}\to\frac{0-1}{0-\frac{1}{5}}=5$$
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