<s>[INST] Limit of $\sqrt{4x+2} -2x$ as x approaches infinity $\hspace{1cm} \displaystyle\lim_{x\to\infty} \left(\sqrt{4x+2}-2x\right) $
I have trouble finding answer to this one, I got stuck. Correct answer seems to be - infinity.
If I try to multiply the algebraic by a ... conjugate, I get:
$(\sqrt{4x+2}-2x) \cdot \frac{\sqrt{4x+2}+2x}{\sqrt{4x+2}+2x} = \frac{4x+2 -4x^2}{\sqrt{4x+2}+2x}  =  \frac {2(2x + 1 - 2x^2)}{\sqrt{4x+2} +2x}$
Dunno what to do next, any suggestions?
 [/INST] Multiply the top and bottom by $\frac 1x$ to your very last expression. You will get $$\frac {2(2x + 1 - 2x^2)}{\sqrt{4x+2} +2x} \frac{\frac 1x}{\frac 1x}=\frac{2(2+\frac 1x - 2x)}{\sqrt{\frac 4x+\frac 2{x^2}}+2}$$
Now take the limit as $x \to \infty$. Your limit will be $-\infty$.
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