Find $\lim_{n\to \infty} \int_n^{n+1} {\sin x \over x} dx$ 
Find $\lim_{n\to \infty} \int_n^{n+1} {\sin x \over x}  dx$

I thought about defining $\space F(x) = \int_0^x {\sin t \over t}  dt \space$ and then the limit is $\space \lim_{n\to \infty} (F(n+1) - F(n))$, so the answer is 0? It doesn't seem ok
 A: EDIT: Similar to Leg's answer
It is known: $ -1 \le \sin(x) \le 1 \ \forall \ x \in \mathbb{R}  $ 
$ \Rightarrow  \int_n^{n+1} {\sin x \over x}  dx \le \ \int_n^{n+1} {1 \over x}  dx = \ln(n+1) - \ln(n) = \ln(1+\frac{1}{n}) \xrightarrow{n \to \infty} 0 $
$ \Rightarrow  \int_n^{n+1} {\sin x \over x}  dx \ge \ -\int_n^{n+1} {1 \over x}  dx = -\ln(n+1) + \ln(n) = - \ln(1-\frac{1}{n}) \xrightarrow{n \to \infty} 0 $
Hence: $\lim\limits_{n\to \infty} \int_n^{n+1} {\sin x \over x} dx = 0  $
A: We have
$$\left \vert \int_n^{n+1} \dfrac{\sin(x)}xdx\right \vert \leq \int_n^{n+1} \left \vert \dfrac{\sin(x)}x\right \vert dx \leq \int_n^{n+1}\dfrac{dx}x = \log\left(1+\dfrac1n\right)$$
Hence, the integral converges to $0$.
A: The function ${\sin x} \over {x}$ is integrable on $(0,\infty)$. This means that the sum
$$ \sum_{n=0}^\infty \int _n ^{n+1} \frac{\sin x}{x} dx =  \int _0 ^{\infty} \frac{\sin x}{x} dx $$ converges. And so, the $n^{th}$ term of this series must go to zero (by the $n^{th}$ term test).
A: You can use the mean value theorem for integrals.
$$\int_{n}^{n+1}{\frac{\sin x}{x}}\,dx=\frac{\sin \varepsilon}{\varepsilon}\; \quad\text{where} \;\varepsilon\in [n;n+1]$$
If $n\to\infty\;$ then $\; \varepsilon\to\infty \;$. Hence $$\lim_{n\to\infty}\int_{n}^{n+1}{\frac{\sin x}{x}}\,dx=\lim_{\epsilon\to\infty}\frac{\sin \varepsilon}{\varepsilon}=0$$
