$\lim_{n\rightarrow \infty} n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})$ $\lim_{n\rightarrow \infty} n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})$, $\forall \alpha \in R$
I can change the form of this limit saying that
$n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})=n^\alpha (\sqrt[5]{n^2(1+\frac{1}{n})}-\sqrt[5]{n^2(1+\frac{2}{n}+\frac{1}{n^2})})=n^{\alpha+\frac{2}{5}} (\sqrt[5]{1+\frac{1}{n}}-\sqrt[5]{1+\frac{2}{n}+\frac{1}{n^2}})$
Then 
$\lim_{n\rightarrow \infty} n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})= \begin{cases}+ \infty & \alpha>-\frac{2}{5}\\
2 & \alpha=-\frac{2}{5}\\0 & \alpha<-\frac{2}{5}\end{cases}$
Is it right?
 A: It's simpler if you consider the limit you get by (formally) substituting $n=1/x$:
$$
\lim_{x\to0^+}\frac{1}{x^{\alpha}}\frac{\sqrt[5]{1+x}-\sqrt[5]{1+2x+x^2}}{\sqrt[5]{x^2}}
$$
If this limit exists (finite or infinite), then it's the same as the limit of your sequence. Now use Taylor:
$$
\frac{1+\dfrac{1}{5}x-1-\dfrac{2}{5}x+o(x)}{x^{\alpha+2/5}}=
\frac{-\dfrac{1}{5}+o(1)}{x^{\alpha-3/5}}
$$
Now it should be easy.
A: By binomial expansion


*

*$\sqrt[5]{n^2+n}=\sqrt[5]{n^2}\sqrt[5]{1+1/n}=\sqrt[5]{n^2}(1+\frac1{5n}+o(1/n))$

*$\sqrt[5]{n^2+2n+1}=\sqrt[5]{n^2}\sqrt[5]{1+2/n+1/n^2}=\sqrt[5]{n^2}(1+\frac2{5n}+o(1/n))$


then
$$n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})=-\frac{n^\alpha\sqrt[5]{n^2}}{5n}+o\left(\frac{n^\alpha\sqrt[5]{n^2}}{n}\right)=-\frac15 n^{\alpha-\frac35}+o\left(n^{\alpha-\frac35}\right)$$
therefore
$$\lim_{n\rightarrow \infty} n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})= \begin{cases}
- \infty & \alpha>\frac{3}{5}\\
-\frac15 & \alpha=\frac{3}{5}\\
0 & \alpha<\frac{3}{5}\end{cases}$$
