how to find this limit : lim x to infinity how can I find this limit which is become infinite
$\lim _{x\to \infty }\left(x(\sqrt{x^2+1}-x)\right)$
can I use conjugate method ?
that what I'm doing until now
$= x\left(\frac{\sqrt{x^2+1}-x}{1}\cdot \:\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}\right)\:$
$= x\left(\frac{1}{\sqrt{x^2+1}+x}\right)$
$= \left(\frac{x}{\left(\sqrt{\frac{x^2}{x^2}+\frac{1}{x^2}}+\frac{x}{x}\right)}\right)\:$
$= \:\left(\frac{x}{\left(\sqrt{1+\frac{1}{x^2}}+1\right)}\right)\:$

For the edited and corrected question, i.e. the limit at $\infty$ of 
$$
x\left(\sqrt{x^2+1} - x\right)
$$
indeed you can deal with this multiplying by the conjugate (cf. jeantheron's answer). However, I would advocate for a more systematic approach, simple enough and highly generalizable. Assuming you have heard or will soon hear about Taylor series, you can do the following (detailing a lot each step):
$$\begin{align}
x\left(\sqrt{x^2+1} - x\right)
&= 
x\left(x\sqrt{1+\frac{1}{x^2}} - x\right)
= 
x\left(x\left(1+\frac{1}{2x^2}+o\left(\frac{1}{x^2}\right)\right) - x\right)
\\
&= 
x\left(x+\frac{1}{2x}+o\left(\frac{1}{x}\right) - x\right)
= 
x\left(\frac{1}{2x}+o\left(\frac{1}{x}\right)\right)
\\
&= 
\frac{1}{2}+o\left(1\right)
\end{align}$$
using that $(1+t)^\alpha = 1+\alpha t+o(t)$ when $t\to0$.
