Limit of a sequence defined as definite integral What is the limit of $$a_n=\int_n^{n+\sqrt{n}}\frac{\sin x}{x}\ dx\ ?$$
What's the key thing to do here?
 A: Apply the mean value theorem for integrals.  Since the integrand is continuous on any interval $[n,n+\sqrt{n}]$ there exists a number $\xi_n$ between $n$ and $n + \sqrt{n}$ such that
$$0\leq\Big{|}\int_n^{n+\sqrt{n}}\frac{\sin x}{x}dx\Big{|}=\frac{|\sin \xi_n|}{\xi_n}\sqrt{n}\leq \frac{1}{\sqrt{n}}\rightarrow0$$
A: Hint: Integrate by parts, using $u=\frac{1}{x}$ and $dv=\sin x\,dx$. Estimate the integral that remains. 
A: Subbing first $u=x-n$ and then $\displaystyle v=\frac{u}{\sqrt{n}}$,
the initial integral is $\displaystyle \int_0^1\frac{\sin(v\sqrt{n}+n)}{v+\sqrt{n}}$
Taking limits under the integral sign yields $0$ as an answer.

In order to justify the limit/integral permutation, note that $$|\frac{\sin(v\sqrt{n}+n)}{v+\sqrt{n}}| \leq\frac{1}{\sqrt{n}} $$
which grants uniform convergence.
A: 
Recall that if $\int_a^\infty f$ exists then 
  $$\lim_{A\to\infty}\int_{A}^\infty f=0$$

The integral
$$\int_0^\infty \frac{\sin x}{x}dx$$
exists (and equal $\frac\pi2$) so
$$a_n=\int_n^\infty \frac{\sin x}{x}dx-\int_{n+\sqrt n}^\infty \frac{\sin x}{x}dx\xrightarrow{n\to\infty}0$$
