How to evaluate $\lim _{n\to \infty }\:\int _{1/(n+1)}^{1/n}\:\frac{\sin\left(x\right)}{x^3}\:dx$? We have to evaluate the following limit: 
$$\lim _{n\to \infty }\:\int _{\frac{1}{n+1}}^{\frac{1}{n}}\:\frac{\sin\left(x\right)}{x^3}\:dx,\:n\in \mathbb{N}$$

First step I wrote that
$\int _{\frac{1}{n+1}}^{\frac{1}{n}}\:\frac{\sin\left(x\right)}{x^3}\:dx\:=\:\frac{1}{n\left(n+1\right)}\cdot f\left(c\right)$ , where $c\in \left(\frac{1}{n+1},\frac{1}{n}\right)$.
I have no ideea how can I evaluate:
$\lim _{n\to \infty }\frac{\:f\left(c\right)}{n\left(n+1\right)}$
P.S: After, I want to see at you another method of solving.Thanks in advance!
 A: Heuristically, $\sin x \approx. x$ for small $x$.  Thus, 
$$\int_{1/(n+1)}^{1/n} \frac{\sin x }{x^3}dx\approx. \int_{1/(n+1)}^{1/n} \left(\frac{1}{x^2} \right)dx\ =1$$
We can make this argument rigorous by writing 
$$\frac{\sin x}{x^3}=x^{-2}+\sum_{k=1}^{\infty} \frac{x^{2k-2}}{(2k+1)!}$$
whereupon
$$\int_{1/(n+1)}^{1/n} \frac{\sin x }{x^3}dx=1+\sum_{k=1}^{\infty} \frac{(\frac{1}{n+1})^{2k-1}-(\frac{1}{n})^{2k-1}}{(2k+1)!(2k-1)}$$
The limit of the series vanishes as $n \to \infty$ and we recover the result we obtained heuristically!
A: Let's generalize. Why? Because this kind of limit doesn't really depend on power series or anything fancy. Suppose we have a continuous $f$ on $(0,1)$ with $\lim_{x\to 0^+}f(x) = 1.$ Then
$$\int_{1/(n+1)}^{1/n}\frac{f(x)}{x^2}\,dx \to 1$$
as $n \to \infty.$ (In our problem we have $f(x) = (\sin x)/x.$) Proof: Let $\epsilon>0.$ Choose $\delta > 0$ such that $1-\epsilon< f(x) < 1 +\epsilon$ for $x\in (0,\delta).$ Then $n>1/\delta$ implies
$$(1-\epsilon)\int_{1/(n+1)}^{1/n}\frac{1}{x^2}\,dx < \int_{1/(n+1)}^{1/n}\frac{f(x)}{x^2}\,dx <(1+\epsilon)\int_{1/(n+1)}^{1/n}\frac{1}{x^2}\,dx.$$As we know, the integrals on the left and right equal $1$ for all $n,$ and this gives the result.
A: HINT: Maybe using this simple inequality and then squeeze it? $$x-\frac{x^3}{6}\le \sin(x) \le x
 \ , x\ge0$$
The limit is $1$.
A: Hint: Note that near $x=0$, $\sin(x)$ is monotonically increasing, therefore
$$
\underbrace{\frac1{n(n+1)}}_{\text{interval width}}\cdot\underbrace{\frac{\sin(1/(n+1))}{1/n^3}}_{\text{integrand min}}
\le\int_{1/(n+1)}^{1/n}\frac{\sin(x)}{x^3}\,\mathrm{d}x
\le\underbrace{\frac1{n(n+1)}}_{\text{interval width}}\cdot\underbrace{\frac{\sin(1/n)}{1/(n+1)^3}}_{\text{integrand max}}
$$
Rearrange a bit to get
$$
\frac{n^3}{n(n+1)^2}\frac{\sin(1/(n+1))}{1/(n+1)}
\le\int_{1/(n+1)}^{1/n}\frac{\sin(x)}{x^3}\,\mathrm{d}x
\le\frac{(n+1)^3}{n^2(n+1)}\frac{\sin(1/n)}{1/n}
$$
Now take limits and apply $\lim\limits_{x\to0}\frac{\sin(x)}x=1$ and the Squeeze Theorem.
A: Hint: $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$
