limit with $\arctan$ I have to find the limit 
and want ask about a hint:
$$\lim_{n \to \infty} n^{\frac{3}{2}}[\arctan((n+1)^{\frac{1}{2}})- \arctan(n^{\frac{1}{2}})]$$
I dont have idea what to do. Derivatives and L'Hôpital's rule are so hard

Write
$$\tan \left(\arctan\left( (n+1)^{1\over2}\right) - \arctan\left( n^{1\over2}\right) \right) = \frac{\sqrt{n+1}-\sqrt{n}}{1+\sqrt{n(n+1)}} = \frac{\sqrt{n}\left(\sqrt{1+{1\over n}}-1\right)}{1+\sqrt{n(n+1)}}$$
Thus 
$$\arctan\left( (n+1)^{1\over2}\right) - \arctan\left( n^{1\over2}\right) = \arctan \left( \frac{\sqrt{n}\left(\sqrt{1+{1\over n}}-1\right)}{1+\sqrt{n(n+1)}}\right) $$
And
$$ \frac{\sqrt{n}\left(\sqrt{1+{1\over n}}-1\right)}{1+\sqrt{n(n+1)}} \sim \frac{n^{1\over 2} \left( 1 + \frac{1}{2n}  + \mathcal{O}\left(\frac{1}{n^2} \right) -1\right)}{n} \sim \frac{1}{2n^{3\over2}}$$
So, since $\arctan x \underset{x \rightarrow 0}{\sim} x$,
$$ \arctan \left( \frac{\sqrt{n}\left(\sqrt{1+{1\over n}}-1\right)}{1+\sqrt{n(n+1)}}\right) \sim \frac{1}{2n^{3\over2}} $$
And your limit is $1 \over 2$.
