Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find the following limit: $$ \lim_{x\to \infty} \sqrt{x^2+x}-\sqrt{x^2-x} $$

I tried to simplify using conjugation. This gave me the following: $$ \lim_{x\to \infty} \frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}} $$

When I plug in the $\infty$, I'm left with $ \frac{\infty}{\infty} $. Did I mess up somewhere, or does the limit not exist?

share|improve this question
    
Since you have $\infty/\infty,$ looks like a job for $l'Hopital.$ Have you tried it? –  Igor Rivin Dec 7 '13 at 19:13
    
I'm aware of the rule, but this is part of review for earlier units in my Calculus class. I don't think my professor wants us to use that on this particular question. –  Hoonsel Dec 7 '13 at 19:21
    

3 Answers 3

up vote 2 down vote accepted

You are almost there:

$$\frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}} = \frac{2x}{x \sqrt{1+\frac{1}{x}}+x\sqrt{1-\frac{1}{x}}}$$

so...

share|improve this answer

Hint:

$$ax^2+bx+c\approx ax^2, ~~x\to\infty$$ so if $x\to+\infty$ then $\sqrt{x^2+x}\approx\sqrt{x^2}=|x|=x$ and if $x\to-\infty$ then $\sqrt{x^2+x}\approx\sqrt{x^2}=|x|=-x$

share|improve this answer
1  
I always like this intuitive approach! +1 –  amWhy Dec 7 '13 at 22:47
    
This nice hint maskes calculus much simpler!+1 صباح الخير أخي قل لي هل تجيد العربية؟ –  Sami Ben Romdhane Dec 8 '13 at 7:03
    
@SamiBenRomdhane: I know some words up to the sharia. Not much. ;-) –  B. S. Dec 8 '13 at 8:03
    
@B.S. Of course you can ask. And I'll try to help in any way I can! –  amWhy Dec 8 '13 at 13:22
    
What was the question you asked? Is he still angry with you about the question? Or are you afraid to bring it up again because of his anger when you asked the question? Did he give you an answer? –  amWhy Dec 8 '13 at 13:39

For $x\ge 1$ we have $$ \sqrt{x^2+x}-\sqrt{x^2-x}=\frac{(x^2+x)-(x^2-x)}{\sqrt{x^2+x}+\sqrt{x^2-x}}=\frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}}=\frac{2}{\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}. $$ It follows that $$ \lim_{x\to\infty}(\sqrt{x^2+x}-\sqrt{x^2-x})=\lim_{x\to\infty}\frac{2}{\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}=1. $$

share|improve this answer

Your Answer

 
discard

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.