Technique to evaluate limits involving a generic continuous function? I have to prove that if $f$ is a continuous function, then:
$$\lim\limits_{h \to 0^+}{\int\limits_{-1}^1{\frac{h}{h^2+x^2}}f(x)\:dx}=\pi f(0)$$
So far what I've tried is noting that $f$ is bounded on $[-1,1]$ so let $M=\sup\limits_{x\in{[-1,1]}}{f(x)}$, and $m=\inf\limits_{x\in{[-1,1]}}{f(x)}$. Then maybe we can use the squeeze theorem somehow to prove the result, but it didn't really lead me anywhere. 
Does anyone have any hints? Are there any techniques to solving problems similar to this-- ones with generic $f$'s under the integrand?

From this manipulation
$$\int_{-1}^1 \frac{h}{h^2+x^2}f(x)dx = \int_{-1}^1 \frac{1}{1+\frac{x^2}{h^2}}f(x)\frac{dx}{h}$$
the substitution $z = \frac{x}{h}$ becomes obvious.
$$ = \int_{-\frac{1}{h}}^{\frac{1}{h}} \frac{f(hz)}{1+z^2}dz$$
Then use dominated convergence to move the limit into the integral
$$ \lim_{h\to 0^+} \int_{-\frac{1}{h}}^{\frac{1}{h}} \frac{f(hz)}{1+z^2}dz = \int_{-\infty}^\infty \frac{f(0)}{1+z^2}dz = f(0)\arctan(z)\Bigr|_{-\infty}^\infty = \pi f(0)$$
Generally the trick is to absorb the limit variable into the $dx$ via some sort of substitution. For example in the integral
$$\lim_{n\to\infty} n\int_0^1 f(x)x^ndx$$
the trick is to use the substitution $u = x^{n+1}$ so that $du \sim x^ndx$
$$= \lim_{n\to \infty} \frac{n}{n+1} \int_0^1 f\left(u^{\frac{1}{n+1}}\right) du = \int_0^1 f(1)du = f(1)$$
