TITLE: Evaluate the limit $\lim_{x\to \infty}( \sqrt{4x^2+x}-2x)$
QUESTION [4 upvotes]: Evaluate :$$\lim_{x\to \infty} (\sqrt{4x^2+x}-2x)$$

$$\lim_{x\to \infty} (\sqrt{4x^2+x}-2x)=\lim_{x\to \infty} \left[(\sqrt{4x^2+x}-2x)\frac{\sqrt{4x^2+x}+2x}{\sqrt{4x^2+x}+2x}\right]=\lim_{x\to \infty}\frac{{4x^2+x}-4x^2}{\sqrt{4x^2+x}+2x}=\lim_{x\to \infty}\frac{x}{\sqrt{4x^2+x}+2x}$$
Using L'Hôpital  $$\lim_{x\to \infty}\frac{1}{\frac{8x+1}{\sqrt{4x^2+x}}+2}$$
What should I do next?

REPLY [3 votes]: With the substitution $x=1/t$ (under the unrestrictive condition that $x>0$) you get
$$
\lim_{x\to \infty}(\sqrt{4x^2+x}-2x)=
\lim_{t\to0^+}\left(\sqrt{\frac{4}{t^2}+\frac{1}{t}}-\frac{2}{t}\right)=
\lim_{t\to0^+}\frac{\sqrt{4+t}-2}{t}
$$
which is the derivative at $0$ of $f(t)=\sqrt{4+t}$; since
$$
f'(t)=\frac{1}{2\sqrt{4+t}}
$$
you have
$$
f'(0)=\frac{1}{4}
$$
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