Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I tried all I know and I always get to $\infty$, Wolfram Alpha says $\frac{3}{2}$. How should I simplify it?

$$\lim\limits_{x \to \infty}\sqrt{(x^2+3x+4)}-x$$

I tried multiplying by its conjugate, taking the squared root out of the limit, dividing everything by $\sqrt{x^2}$, etc.

Obs.: Without using l'Hôpital's.

share|cite|improve this question
up vote 5 down vote accepted

Note that \begin{align} \sqrt{x^2+3x-4} - x & = \left(\sqrt{x^2+3x-4} - x \right) \times \dfrac{\sqrt{x^2+3x-4} + x}{\sqrt{x^2+3x-4} + x}\\ & = \dfrac{(\sqrt{x^2+3x-4} - x)(\sqrt{x^2+3x-4} + x)}{\sqrt{x^2+3x-4} + x}\\ & = \dfrac{x^2+3x-4-x^2}{\sqrt{x^2+3x-4} + x} = \dfrac{3x-4}{\sqrt{x^2+3x-4} + x}\\ & = \dfrac{3-4/x}{\sqrt{1+3/x-4/x^2} + 1} \end{align} Now we get \begin{align} \lim_{x \to \infty}\sqrt{x^2+3x-4} - x & = \lim_{x \to \infty} \dfrac{3-4/x}{\sqrt{1+3/x-4/x^2} + 1}\\ & = \dfrac{3-\lim_{x \to \infty} 4/x}{1 + \lim_{x \to \infty} \sqrt{1+3/x-4/x^2} } = \dfrac{3}{1+1}\\ & = \dfrac32 \end{align}

share|cite|improve this answer
I'm really sorry for taking your time friend, I realized that I was working with $+1$ instead of $+x$ all the time. Thank you for your kind answer and sorry for this. – Luan Cristian Thums Jun 9 '13 at 20:41

Intuitively you can see this as follows:
Write $x^2+3x+4$ as $\left(x+\frac32\right)^2+\frac74$. For $x$ large this quantity is almost the same as $\left(x+\frac32\right)^2$. Therefore for $x$ large $\sqrt{x^2+3x+4}-x\sim\sqrt{\left(x+\frac32\right)^2}-x=\frac32$

share|cite|improve this answer

$$ \lim_{x \rightarrow \infty} \left(x^2 + 3x + 4\right)^{ \frac{1}{2}} - x $$ $$ = \lim_{x \rightarrow \infty} \left(x^2\left(1+\frac{3}{x}+\frac{4}{x^2}\right)\right)^{ \frac{1}{2}} - x $$ $$ = \lim_{x \rightarrow \infty} x\left(1+\frac{3}{x}+\frac{4}{x^2}\right)^{ \frac{1}{2}} - x $$

Then by Taylor expansion, we get that

$$ = \lim_{x \rightarrow \infty} x\left(1+\frac{1}{2}\left(\frac{3}{x}+\frac{4}{x^2}\right)+\operatorname{o}\left(\frac{1}{x^2}\right)\right) - x $$

$$ = \lim_{x \rightarrow \infty} \frac{3}{2} + \operatorname{o}\left(\frac{1}{x}\right) = \frac{3}{2} $$

as required.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.