4

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?

Martin Sleziak
  • 53,687
gbox
  • 12,867
  • @ΘΣΦ GenSan Why do you keep digging up old questions with minor edits? These questions do not need to be bumped, which they are now. – TMM Feb 01 '17 at 16:04

3 Answers3

10

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 $x$

Empty
  • 13,012
10

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

Lutz Lehmann
  • 126,666
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 $$ f'(0)=\frac{1}{4} $$

egreg
  • 238,574
  • 3
    That is a little bit of cheating. You want to calculate the right limit while using the derivative of the expression. But for calculating the derivative you have to calculate the limit. – Richard Feb 18 '16 at 08:19
  • 1
    @Richard I use to say that with derivatives we know how to compute many limits. We have a very powerful machine that, in many cases, allows us to avoid complicated tricks. For instance, since $f(x)=x^a=\exp(a\log x)$, we can say that $f'(x)=\frac{a}{x}\exp(a\log x)=x^{a-1}$ (for $x>0$), and this avoids all “rationalization” tricks when $a=p/q$ is rational. Just apply the chain rule and the derivatives of $\exp$ and $\log$. – egreg Feb 18 '16 at 08:33