Question: Show that 
$$ 
\int\_0^{\infty} \frac{\sin \pi \omega \sin x\omega}{1-\omega^2}d\omega=
\begin{cases}
\frac{\pi}{2}\sin x,&\mbox{ if } 0\leq x\leq\pi\\
\quad\\
0,&\mbox{ if } x\geq\pi
\end{cases}
$$

Here is what I did 
$$
f(x)=
\begin{cases}
\sin x,&\mbox{ if } 0\leq x\leq\pi\\
\quad\\
0,&\mbox{ if } x\geq\pi
\end{cases}
$$
Then I tried to compute the integral

$$
\int\_0^{\infty}[A(\omega)\cos(\omega x)+B(\omega)\sin(\omega x)]d\omega
$$

But I never reach my target. What is wrong?

Answer: 
If you agree that

$$\lim\_{x\to0^+}{1\over x}=\lim\_{x\to0}{1\over |x|}$$

then, using an appropriate "multiplication" theorem for limits of the form $\lim\_{x\to a}(f(x)g(x))=\lim\_{x\to a}(f(x))\lim\_{x\to a}(g(x))$, we have

$$\lim\_{x\to0}{1\over x^2}=\lim\_{x\to0}{1\over|x|^2}=\left(\lim\_{x\to0}{1\over|x|} \right)^2=\left(\lim\_{x\to0^+}{1\over x} \right)^2=\infty$$

**Added later** (at OP's request): All you really need here is a theorem that says 

$$\lim\_{x\to a}f(x)=\lim\_{x\to a}g(x)=\infty\implies \lim\_{x\to a}(f(x)g(x))=\infty$$

But this follows easily from the definition of $\infty$ as a limit: $\lim\_{x\to a}f(x)=\infty$ if for all $M\gt0$ (no matter how large) there exists a $\delta\gt0$ such that $|x-a|\lt\delta\implies f(x)\gt M$. Here's the proof: Let $M\gt0$, and let $M=M\_1M\_2$ with $M\_1,M\_2\gt0$. Since $\lim\_{x\to a}f(x)=\infty$, there is a $\delta\_1\gt0$ such that $|x-a|\lt\delta\_1\implies f(x)\gt M\_1$. Likewise, since $\lim\_{x\to a}g(x)=\infty$, there is a $\delta\_2\gt0$ such that $|x-a|\lt\delta\_2\implies g(x)\gt M\_2$. Now let $\delta=\min(\delta\_1,\delta\_2)$. Then $|x-a|\lt\delta$ implies $f(x)\gt M\_1$ and $g(x)\gt M\_2$, hence $f(x)g(x)\gt M\_1M\_2=M$, and thus $\lim\_{x\to a}(f(x)g(x))=\infty$.

Note: These basic "arithmetic" theorems for limits *do* require the formal (epsilon-delta) definitions for their proofs. But once you have the theorems at your disposal, you can cut way back on the need for the formal definitions. You just have to make sure you're applying the theorems correctly.
If $|x| < 1$,
then
$\frac1{x^2}
> |\frac1{x}|$,
so the $\delta$ you use
to show
$\frac1{x} \to \infty$
will also work for
$\frac1{x^2}$.

For a proof that
$\lim\_{x \to a}f(x) = \infty$,
you need to show that,
for any $V > 0$,
there is a $c(V) > 0$
such that
$|x-a| < c(V)$
implies
$f(x) > V$.

My comment above shows that
a $c(V)$ that works for
$\frac1{x}$
will works for
$\frac1{x^2}$,
but that is not necessary
in the proof.