TITLE: Find $\lim_{n\to \infty} \int_n^{n+1} {\sin x \over x} dx$
QUESTION [6 upvotes]: 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

REPLY [3 votes]: 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  $