Find an equivalent of this function, a) $f$ continuous on $[0,1]$ such that $f(x)>0$.  Find an equivalent of $$h(\epsilon) = \int_0^1 \frac {f(x)}{x^2 + \epsilon^2}dx$$
when $\epsilon$ goes to zero and when $\epsilon$ goes to infinity.
b) Assume $f$ is differentiable at $0$, that $f(0) = 0$ and $f'(0) \ne 0$. Find an equivalent of $h(\epsilon)$, when $\epsilon$ goes to zero.
Edit:  I think I am satisfied with my answer to part(a).  So, if anyone can help with part(b), that would be great :-)
Thanks,
 A: For part (a), we have (by the continuity of $f$), that there exists some $\delta \in (0,1)$ such that $f(x) > \frac{1}{2} f(0)$ for all $0 \le x \le \delta$.  For this $\delta$, 
$$\int_0^\delta \frac{f(x)}{x^2+\epsilon^2} \; dx \ge \frac{1}{2}f(0)\int_0^\delta \frac{1}{x^2+\epsilon^2}\;dx = \frac{1}{2\epsilon}f(0)\arctan\left(\frac{\delta}{\epsilon}\right)$$
As $\epsilon \to 0$, the righthand expression goes to $+\infty$, so the integral from $0$ to $\delta$ blows up.  On the other hand, 
$$\left|\int_\delta^1 \frac{f(x)}{x^2+\epsilon^2}\;dx\right| \le \int_\delta^1 \frac{|f(x)|}{x^2+\epsilon^2}\;dx \le \int_\delta^1 \frac{|f(x)|}{x^2}\;dx < \infty$$
so the contribution to the rest of the integral on $[\delta, 1]$ is finite, and so the integral blows up.
For part (b), suppose for simplicity that $f'(0) > 0$ (the argument goes through almost identically for $f'(0) < 0$).  Then, by the definition of derivative, $f(x) = f'(0)x + o(x)$ near $0$, so in particular for some $\delta \in(0,1)$, $f(x) > \frac{f'(0)}{2}x$, and so
$$\int_0^\delta \frac{f(x)}{x^2+\epsilon^2}\;dx \ge \frac{f'(0)}{2}\int_0^\delta \frac{x}{x^2+\epsilon^2}\;dx = \frac{f'(0)}{4}\int_{\epsilon^2}^{\delta^2+\epsilon^2} \frac{du}{u}$$
evaluating this integral and using reasoning similar to what was used in part (a), you'll find that the integral blows up to $+\infty$ (or $-\infty$, if $f'(0) < 0$).

If you interpret 'find an equivalent' as 'find an asymptotic expression,' then I believe (the important parts of) my answer can still be used: you'll just have to modify my bounds to (1) be two sided, and (2) be 'shrinkable' to $1$ (so, say for part (a), $(1-\eta)f(0) < f(x) < (1+\eta)f(0)$, with $\eta$ some arbitrary parameter).
A: It is interesting to note that 
$$\lim_{\epsilon\to0}\epsilon\,\int_0^1\frac{f(x)}{x^2+\epsilon^2}\,dx=\frac{\pi}{2}\,f(0)$$
We can show this as follows. First, we have
$$\begin{align}
\lim_{\epsilon\to0}\epsilon\,\int_0^1\frac{1}{x^2+\epsilon^2}\,dx&=\lim_{\epsilon\to0}\arctan(1/\epsilon)\\\\
&=\pi/2
\end{align}$$
Next, by continuity of $f$ we have that given $\nu>0$, there exists a $\delta>0$ such that $|x|<\delta\implies |f(x)-f(0)|<\nu/\pi$.  Denoting $M=||f(x)||_{\infty}$, we have
$$\begin{align}
\left|\int_0^1(f(x)-f(0))\frac{\epsilon}{x^2+\epsilon^2}\,dx\right|&\le\int_0^{\delta}|f(x)-f(0)|\frac{\epsilon}{x^2+\epsilon^2}\,dx+\int_{\delta}^{1}|f(x)-f(0)|\frac{\epsilon}{x^2+\epsilon^2}\,dx \\\\
&\le\nu/2+\int_{\delta}^{1}|f(x)-f(0)|\frac{\epsilon}{x^2+\epsilon^2}\,dx\\\\
&\le\nu/2+\frac{2M\epsilon}{\delta^2}\\\\
&\le \nu
\end{align}$$
if we take $\epsilon<\nu\delta^2/(4M)$.  This gives the desired result
$$\bbox[5px,border:2px solid #C0A000]{\lim_{\epsilon\to0}\epsilon\,\int_0^1\frac{f(x)}{x^2+\epsilon^2}\,dx=\frac{\pi}{2}\,f(0)}$$
which is equivalent to the statement the in the sense of generalized functions
$$\bbox[5px,border:2px solid #C0A000]{\lim_{\epsilon\to0}\frac{\epsilon}{x^2+\epsilon^2}=\frac{\pi}{2}\delta(x)}$$

For the problem at hand, we have
$$\lim_{\epsilon\to 0}\int_0^1\frac{f(x)}{x^2+\epsilon^2}dx$$
and therefore, we can view the integral in the sense of generalized functions with
$$\int_0^1\frac{f(x)}{x^2+\epsilon^2}dx\sim \frac{\pi/2}{\epsilon}f(0)$$

NOTE:
Another way to view this problem is as follows.  For $f$ continuous only, we can write as before for $\delta>0$
$$\begin{align}
\int_0^1\frac{f(x)}{x^2+\epsilon^2}dx&=\int_0^{\delta}\frac{f(x)}{x^2+\epsilon^2}dx+\int_{\delta}^1\frac{f(x)}{x^2+\epsilon^2}dx\\\\
&=\frac{1}{\epsilon}\left(\frac{\pi}{2}-\arctan\left(\frac{\epsilon}{\delta}\right)\right)f(\xi_{<})\\\\
&+\left( \arctan \left(\frac{\epsilon}{\delta}\right)-\arctan\left(\epsilon\right)\right)f(\xi_{>})
\end{align}$$
by the mean-value theorem for some $0<\xi_{<}<\delta$ and $\delta < \xi_{>}<1$.  We can expand the arctangent expressions using
$$\arctan(x)=x-\frac13x^3+\frac15x^5+O\left(x^6\right)$$
Then, we have
$$\begin{align}
\int_0^1\frac{f(x)}{x^2+\epsilon^2}dx&=\frac{1}{\epsilon}\frac{\pi}{2}f(\xi_{<})-\frac{1}{\delta}\left(1-\frac13\left(\frac{\epsilon}{\delta}\right)^2+\frac15\left(\frac{\epsilon}{\delta}\right)^4+O\left(\frac{\epsilon}{\delta}\right)^5\right)f(\xi_{<})\\\\
&+\frac{(1-\delta)\epsilon}{\delta}\left(1-\frac13(1+\delta+\delta^2)\left({\epsilon}{\delta}\right)^2+\frac15(1+\delta+\delta^2+\delta^3+\delta^4)\left({\epsilon}{\delta}\right)^4+O\left(\frac{\epsilon}{\delta}\right)^5\right)f(\xi_{>})
\end{align}$$
Now, we see that the expression
$$\int_0^1\frac{f(x)}{x^2+\epsilon^2}dx-\frac{\pi/2}{\epsilon}f(0)$$
makes no sense in the limit as $\epsilon\to 0$, but the expression
$$\frac{\int_0^1\frac{f(x)}{x^2+\epsilon^2}dx}{\frac{\pi/2}{\epsilon}f(0)} \to 0\tag 1$$
Therefore, we see again that 
$$\int_0^1\frac{f(x)}{x^2+\epsilon^2}dx\sim \frac{\pi/2}{\epsilon}f(0)$$
in the sense of $(1)$.  Similarly, we can show that if $f'(0)$ exists, then
$$\int_0^1\frac{f(x)}{x^2+\epsilon^2}dx\sim \frac{\pi/2}{\epsilon}f(0)-\log (\epsilon) f'(0)$$
