Calculate $\lim_{n\rightarrow \infty}n\ \int_0^n \sin^2(\pi x)\left(\frac{1}{(x-1)(x-2)}+\frac{1}{(x+1)(x+2)}\right) dx$ Calculate
$$\lim_{n\rightarrow \infty}n\ \int_0^n \sin^2(\pi x)\left(\frac{1}{(x-1)(x-2)}+\frac{1}{(x+1)(x+2)}\right) dx$$
Maybe this is too hard for me, Any suggestions please?
EDIT.
Since
$$\lim_{n\rightarrow \infty}\int_0^n \sin^2(\pi x)\left(\frac{1}{(x-1)(x-2)}+\frac{1}{(x+1)(x+2)}\right) dx=0$$
I used L'Hospital's rule on
$$\lim_{t\rightarrow \infty}\frac{\int_0^t \sin^2(\pi x)\left(\frac{1}{(x-1)(x-2)}+\frac{1}{(x+1)(x+2)}\right) dx}{1/t}=$$
$$=\lim_{t\rightarrow \infty}\frac{\sin^2(\pi t)\left(\frac{1}{(t-1)(t-2)}+\frac{1}{(t+1)(t+2)}\right)}{-1/t^2}$$
i.e. 
$$=\lim_{n\rightarrow \infty}\frac{\sin^2(\pi n)\left(\frac{1}{(n-1)(n-2)}+\frac{1}{(n+1)(n+2)}\right)}{-1/n^2}=0$$
by $\sin(\pi n)=0$. Is it right?
 A: No, it isn't.
$$ f(t)= t^2\sin^2(\pi t)\left(\frac{1}{(t-1)(t-2)}+\frac{1}{(t+1)(t+2)}\right) $$
keeps oscillating between $0$ and $2$ for large $t$, so that $\lim_{t\to +\infty} f(t)\,dt $ does not exist.
However, if you perform integration by parts before applying De l'Hopital theorem, i.e. if you exploit the fact that the mean value of $\sin^2(\pi x)$ is just $\frac{1}{2}$, you will see that your limit is exactly $\color{red}{-1}$:
$$\begin{eqnarray*} &&\int_{0}^{n}\sin^2(\pi x)\left(\frac{1}{(x-1)(x-2)}+\frac{1}{(x+1)(x+2)}\right)\,dx\\&=&\frac{n(n^2+2)}{(n^2-1)(n^2-4)}+\int_{0}^{n}\left(\frac{x}{2}+\frac{\sin(2\pi x)}{4\pi}\right)\frac{4x(x^4+4x^2-14)}{(x^2-1)^2(x^2-4)^2}\,dx.\end{eqnarray*} $$
A: Lemma: $n\int_n^{n+1}\frac{\sin^2 (\pi x)}{x}\,dx \to \int_0^1\sin^2 (\pi x)\,dx = 1/2.$
Proof: The integrand lies between $\sin^2 (\pi x)/(n+1)$ and $\sin^2 (\pi x)/n.$ Because $\sin^2 (\pi x)$ has period $1,$ the expression of interest lies between
$$ \frac{n}{n+1}\int_0^1\sin^2 (\pi x)\,dx, \int_0^1\sin^2 (\pi x)\,dx,$$
giving the lemma.
To find the limit in the problem, note
$$\frac{1}{(x-1)(x-2)}= \frac{1}{x-2}-\frac{1}{x-1},\,\,\,\,
\frac{1}{(x+1)(x+2)}= \frac{1}{x+1}-\frac{1}{x+2}.$$
Integrate $\sin^2 (\pi x)$ against each of these fractions and make simple changes of variables. The integral equals
$$\left(\int_1^{n+1} - \int_2^{n+2}+\int_{-2}^{n-2} - \int_{-1}^{n-1}\right) \frac{\sin^2 (\pi x)}{x}\,dx.$$
After cancellation we have
$$\left(\int_1^{2} - \int_{n+1}^{n+2}+\int_{-2}^{-1} - \int_{n-2}^{n-1}\right)\frac{\sin^2 (\pi x)}{x}\,dx.$$
Because the integrand is odd, the first and third integrals cancel. Now multiply by $n$ and take limits using the lemma. We get $-1$ for the answer.
