$ \lim_{x \to \infty} x(\sqrt{2x^2+1}-x\sqrt{2})$ Any ideas fot evaluating:
$$ \lim_{x \to \infty} x(\sqrt{2x^2+1}-x\sqrt{2})$$
thanks.
 A: While one way to proceed here is to write the term 
$\sqrt{2x^2+1}-\sqrt{2}x=\frac{1}{\sqrt{2x^2+1}+\sqrt{2}x}$
a second way is to write
$$\sqrt{2x^2+1}=\sqrt{2}x\left(1+\frac{1}{4x^2}+O\left(\frac{1}{x^4}\right)\right)$$
so that 
$$\begin{align}
x\left(\sqrt{2x^2+1}-\sqrt{2}x\right)&=\sqrt{2}x^2\left(1+\frac{1}{4x^2}+O\left(\frac{1}{x^4}\right)\right)-\sqrt{2}x^2\\\\
&=\frac{\sqrt{2}}{4}+O\left(\frac{1}{x^2}\right)\\\\
&\to \frac{\sqrt{2}}{4}
\end{align}$$
A: HINT:
Multiply by
$$\frac{\sqrt{2x^2+1} + x\sqrt{2}}{\sqrt{2x^2+1} + x\sqrt{2}}$$
to get 
$$x(\sqrt{2x^2+1} - x\sqrt{2})\cdot \frac{\sqrt{2x^2+1} + x\sqrt{2}}{\sqrt{2x^2+1} + x\sqrt{2}}= x\cdot \frac{x}{\sqrt{2x^2+1} + x\sqrt{2}}$$
At this point, you should see that the numerator has degree 1 and (in some sense) the denominator has degree 1 too (square root of a square), as David suggests on the comments, dividing by $x$ should gives you a light (L'Hopital should work too).
A: After rationalization it become
$$=\lim_{x\to\infty}\frac{x}{\sqrt{2x^2+1}+x\sqrt{2}}$$
Now you need to eliminate $x$ from nominator to define answer because function is undefined i.e. $\frac{\infty}{\infty}$ after applying limit. Take common $x$ both from nominator and denominator
$$=\lim_{x\to\infty}\frac{1}{\sqrt{{2}+\frac{1}{x^2}}+{\sqrt{2}}}$$
Now after applying limit it will become
$$=\frac{1}{\sqrt{2+0}+\sqrt{2}}$$
$$=\frac{1}{2\sqrt{2}}$$
