Evaluate the limit $\lim_{x\to \infty}( \sqrt{4x^2+x}-2x)$

Question

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?

Answer 1

Hint :

$\displaystyle\lim_{x\to \infty}\frac{x}{\sqrt{4x^2+x}+2x}=\lim_{x\to \infty}\frac{1}{\sqrt{4+\frac{1}{x}}+2}$ , dividing numerator and denominator by $x$

Answer 2

Do not use the sledge hammer l'Hopital. Just cancel $x$ to $$ \frac1{\sqrt{4+\frac1x}+2} $$

Answer 3

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 $$ f'(0)=\frac{1}{4} $$