Calculate the limit help

1. Mar 10, 2009

walker242

1. The problem statement, all variables and given/known data
Calculate the limit of
$$\lim_{x\rightarrow \infty} \frac{\sqrt{x^{2}+5} - x}{\sqrt{x^{2}+2} - x}$$

2. Relevant equations
-

3. The attempt at a solution
Neither multiplying with the conjugate nor trying to break out x helps me, as I'm left with "0/0" in those cases.

2. Mar 10, 2009

lanedance

Re: $\lim_{x\rightarrow\infty} \frac{\sqrt{x^{2}+5} - x}{\sqrt{x^{2}+2} - x}$

do you know l'hopitals rule? could help here i think

3. Mar 10, 2009

walker242

Re: $\lim_{x\rightarrow\infty} \frac{\sqrt{x^{2}+5} - x}{\sqrt{x^{2}+2} - x}$

While I do know l'Hopitals rule, we have not yet covered it in the course. The problem should be solved without using (sadly).

4. Mar 10, 2009

lanedance

Re: $\lim_{x\rightarrow\infty} \frac{\sqrt{x^{2}+5} - x}{\sqrt{x^{2}+2} - x}$

$$\lim_{x\rightarrow \infty} \frac{5 \sqrt{x^{2}+2} + x}{2 \sqrt{x^{2}+5} +x}$$

already looking in better shape, as its not a difference term that is leading to the zero, which ws the tricky bit, so from here I'd try multiplying through by:

$$\lim_{x\rightarrow \infty} \frac{\frac{1}{x}}{\frac{1}{x}}$$

this should change the terms containing x in the numerator & denominator from tending to infinity, to ones tending to zero...

5. Mar 10, 2009

walker242

Re: $\lim_{x\rightarrow\infty} \frac{\sqrt{x^{2}+5} - x}{\sqrt{x^{2}+2} - x}$

$$\lim_{x\rightarrow \infty} \frac{5\sqrt{x^{2}+2} + x}{2\sqrt{x^{2}+5} +x} = \lim_{x\rightarrow\infty} = \frac{5}{2}\frac{x(\sqrt{1+\frac{2}{x}})+1}{x(\sqrt{1+\frac{5}{x}})+1} = \frac{5}{2}\cdot\frac{2}{2} = \frac{5}{2}$$

Cheers!

6. Feb 9, 2012

checkitagain

Re: $\lim_{x\rightarrow\infty} \frac{\sqrt{x^{2}+5} - x}{\sqrt{x^{2}+2} - x}$

There were some significant errors in this (highlighted in the

quote box) post from a user, that I felt one of the corrected

versions should be shown.

7. Feb 9, 2012

micromass

Staff Emeritus
Re: [itex]\lim_{x\rightarrow\infty} \frac{\sqrt{x^{2}+5} - x}{\sqrt{x^{2}+2} - x}[/it

This thread is 2 years old. You've been here long enough to know not to necropost.