<s>[INST] $ \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.
 [/INST] 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).
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