Simplify this limit Suppose $f$ is a continuous function on $[a,b]$. Is there anyway I can simplify this limit? 

$$\lim_{\epsilon \to 0}\int_a^b \dfrac{\epsilon f(t) dt}{(t-x)^2 +\epsilon^2 }$$ 
  where $x \in [a,b]$.

 A: $f$ is continuous on $[a,b]$, so by MVT, there exists a $c \in [a,b]$ such that:
$$\int_a^b \dfrac{\epsilon f(t) dt}{(t-x)^2 +\epsilon^2 } = f(c)\int_a^b \dfrac{\epsilon}{(t-x)^2 +\epsilon^2 }dt$$
But:
$$\int_a^b \dfrac{\epsilon}{(t-x)^2 +\epsilon^2 }dt = \cdots = \arctan(\frac{b-x}{\epsilon}) - \arctan(\frac{a-x}{\epsilon})$$
The limit eventually becomes: $\pi f(c)$ if $x \in ]a,b[$, $\frac{\pi}{2}f(c)$ if $x = a$ or $x = b$, 
A: The integral from $a$ to $b$ breaks into two pieces:  $\int_a^b=\int_a^x+\int_x^b$.  Let's consider the second of these.  Using the change of variables $t-x=\epsilon u$ (as in abel's answer), we have
$$\int_x^b{\epsilon f(t)dt\over(t-x)^2+\epsilon^2}=\int_0^{(b-x)/\epsilon}{f(x+\epsilon u)du\over u^2+1}$$
Now if $x=b$, this integral is clearly $0$ for all $\epsilon\not=0$.  If $b\lt x$, on the other hand, the integral converges to
$$\int_0^\infty{f(x)du\over u^2+1}={\pi\over2}f(x)$$
as $\epsilon\to0$. A similar argument applies to the piece from $a$ to $x$, so overall we get the limit $\pi f(x)$ if $a\lt x\lt b$ and ${\pi\over2}f(x)$ if $x=a$ or $b$.
I hasten to point out that a bit of care actually needs to be taken to formally justify simultaneously letting $(b-x)/\epsilon\to\infty$ and $f(x+\epsilon u)\to f(x)$.  I'm glossing over it partly out of laziness, but I'll flesh it out if the OP requests.
