Calculating two specific limits with Euler's number I got stuck, when I were proving that
$$\lim_{n \to \infty} \frac {\sqrt[2]{(n^2+5)}-n}{\sqrt[2]{(n^2+2)}-n} = \frac {5}{2}$$
$$\lim_{n \to \infty}n(\sqrt[3]{(n^3+n)}-n) = \frac {1}{3}$$
First one I tried to solve like 
$$\lim_{n \to \infty} \frac {\sqrt[2]{(n^2+5)}-n}{\sqrt[2]{(n^2+2)}-n} = \lim_{n \to \infty} \frac {n\sqrt[2]{(1+\frac {5}{n^2})}-n}{n\sqrt[2]{(1+\frac{2}{n^2})}-n}= \lim_{n \to \infty} \frac {\sqrt[2]{(1+\frac {5}{n^2})}-1}{\sqrt[2]{(1+\frac{2}{n^2})}-1}$$
and now I think, that this one sholud go like $$\lim_{n \to \infty} \frac{\frac{5}{n^2}}{\frac{2}{n^2}}=\frac{5}{2} $$
but I have no idea how to prove this. 
In the second one I made 
$$\lim_{n \to \infty}n(\sqrt[3]{(n^3+n)}-n) = \lim_{n \to \infty}n(n\sqrt[3]{(1+\frac{1}{n^2}}-n)= \lim_{n \to \infty}n^2(\sqrt[3]{(1+\frac{1}{n^2}}-1)= \lim_{n \to \infty}n^2(e^{\frac{1}{3n^2}}-1)  $$
And now I do not know what to do next...
I would be really grateful, for any help, or prompt, how to solve these ones (or information, where is the mistake).
 A: $$(a)\;\;\lim_{n\rightarrow \infty}\frac{\sqrt{n^2+5}-n}{\sqrt{n^2+2}-n} =\lim_{n\rightarrow \infty}\frac{\sqrt{n^2+5}-n}{\sqrt{n^2+2}-n}\times \frac{\sqrt{n^2+5}+n}{\sqrt{n^2+5}+n}\times \frac{\sqrt{n^2+2}+n}{\sqrt{n^2+2}+n} $$
So we get $$=\lim_{n\rightarrow \infty}\frac{5}{2}\times \frac{\sqrt{n^2+2}+n}{\sqrt{n^2+5}+n} = \frac{5}{2}\lim_{n\rightarrow \infty}\frac{n(\sqrt{1+\frac{2}{n^2}}+1)}{n(\sqrt{1+\frac{5}{n^2}}+1)}=\frac{5}{2}$$
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$$(b) \lim_{n\rightarrow \infty}n\left[\sqrt[3]{n^3+n}-n\right]$$
Now Put $\displaystyle n=\frac{1}{y}\;,$ When $n\rightarrow \infty\;,$ Then $y\rightarrow 0$
So we get $$\lim_{y\rightarrow 0}\frac{(1+y^2)^{\frac{1}{3}}-1}{y^2}$$
Now Let $(1+y^2)^{\frac{1}{3}}=A$ and $1=B\;,$ Then $A^3-B^3 = 1+y^2-1=y^2$
So we get $$\lim_{y\rightarrow 0}\frac{A-B}{A^3-B^3} = \lim_{y\rightarrow 0}\frac{A-B}{(A-B)(A^2+B^2+AB)} = \lim_{y\rightarrow 0}\frac{1}{A^2+B^2+AB}$$
So we get $$\lim_{y\rightarrow 0}\frac{1}{(1+y^2)^{\frac{2}{3}}+1^2+(1+y^2)^{\frac{1}{3}}} = \frac{1}{3}$$
A: When you arrive at 
$$\lim_{n \to \infty} \frac {\sqrt[2]{(1+\frac {5}{n^2})}-1}{\sqrt[2]{(1+\frac{2}{n^2})}-1},$$
you can continue by
$$\lim_{n \to \infty} \frac {\sqrt[2]{(1+\frac {5}{n^2})}-1}{\sqrt[2]{(1+\frac{2}{n^2})}-1}=\lim_{n \to \infty} \frac {\sqrt[2]{(1+\frac {5}{n^2})}-1}{\sqrt[2]{(1+\frac{2}{n^2})}-1}\cdot
\frac {\sqrt[2]{(1+\frac {5}{n^2})}+1}{\sqrt[2]{(1+\frac{2}{n^2})}+1}\cdot
 \frac {\sqrt[2]{(1+\frac{2}{n^2})}+1}{\sqrt[2]{(1+\frac {5}{n^2})}+1}$$
$$=\lim_{n \to \infty} \frac {   1+\frac {5}{n^2} -1}{ { 1+\frac{2}{n^2}}-1}
\cdot
 \frac {\sqrt[2]{(1+\frac{2}{n^2})}+1}{\sqrt[2]{(1+\frac {5}{n^2})}+1}=\frac 52\lim_{n \to \infty}\frac {\sqrt[2]{(1+\frac{2}{n^2})}+1}{\sqrt[2]{(1+\frac {5}{n^2})}+1},$$
and the last limit should be easy to find (both numerator and denominator converge to a finite non-zero number).
The same trick works for the second limit. Just observe that
$$n(\sqrt[3]{(n^3+n)}-n) = n\frac{\big((n^3+n)^{1/3}-n\big)\big((n^3+n)^{2/3}+(n^3+n)^{1/3}n+n^2\big)}{(n^3+n)^{2/3}+(n^3+n)^{1/3}n+n^2} = n\frac{n^3+n-n^3}{(n^3+n)^{2/3}+(n^3+n)^{1/3}n+n^2} = \frac{1}{(1+1/n^2)^{2/3}+(1+1/n^2)^{1/3}+1}.$$
