How to show $\int_0^\infty\frac{dx}{x\sqrt{1-x^2}}=\pi/2$ How to show that 
$$\int_0^\infty\frac{dx}{x\sqrt{1-x^2}}=\frac{\pi}{2}$$
The problem is that I don't know what is 
$$\lim\limits_{x\to\infty}{\mathrm{arcsec}\ x}$$

Assuming that you intended to write 
$$ 
\int_1^\infty \frac{1}{x\sqrt{x^2-1}}dx=\frac{\pi}{2}
$$
First note that
$$ 
\int_1^\infty \frac{1}{x\sqrt{x^2-1}}dx= \int_1^\infty \frac{x}{x^2\sqrt{x^2-1}}dx
$$
Let $u=\sqrt{x^2-1}$, then $du=\frac{x}{\sqrt{x^2-1}}dx$. So now
$$ 
\int_0^\infty \frac{1}{u^2+1}du
$$
$$=\lim\limits_{\beta\to \infty} \arctan \beta- \lim\limits_{\alpha\to 0^+} \arctan \alpha
$$
$$=\frac{\pi}{2}- 0= \frac{\pi}{2}
$$
As already mentioned, your original integral does not converge.
