# Homework Help: Proving a Limit.

1. Feb 17, 2012

### mtayab1994

1. The problem statement, all variables and given/known data

$$f(x)=\frac{\sqrt{x+2}-\sqrt{2}}{\sqrt{x}}$$

2. Relevant equations

prove that: $$\lim_{x\rightarrow0}f(x)=0$$

3. The attempt at a solution

Alright i let ε>0 and i have to find α>0 (which depends on ε) so that if:

0<l x-a l< α, then l f(x)-L l<ε

Any help guys thank you before hand.

2. Feb 17, 2012

### sunjin09

differentiate both numerator and denominator

Last edited: Feb 17, 2012
3. Feb 17, 2012

### mtayab1994

I haven't learned differentials yet, is there any other way to do it?

4. Feb 17, 2012

### sunjin09

If you let $u=\sqrt{x+2}-\sqrt{2}$, you'll find
$f=\frac{1}{\sqrt{1+2\sqrt{2}/u}}$
the rest is obvious

5. Feb 19, 2012

### mtayab1994

Yes but i have to prove that its true using the limit rule:

For every real ε > 0, there exists a real δ > 0 such that for all real x, 0 < | x − p | < δ implies | f(x) − L | < ε.

6. Feb 19, 2012

### mtayab1994

I multiplied the numerator and the denominator by the conjugate of the numerator and at the end i got:

$$f=\frac{\sqrt{x}}{\sqrt{x-2}+\sqrt{2}}$$

and i substitute with the zero and i get that the limit as x approaches 0 of f(x) is 0 is that correct?

7. Feb 19, 2012

### vela

Staff Emeritus
Your idea is correct, but the numerator should be just x, not the square root of x.

8. Feb 19, 2012

### mtayab1994

alright is that all that needs to be done?

9. Feb 19, 2012

### Staff: Mentor

Your denominator is incorrect. If you take the limit as x approaches 0, the first radical in the denominator is undefined. Check your work.

10. Feb 20, 2012

### mtayab1994

sorry i wanted to put √(x+2)

11. Feb 20, 2012

### SammyS

Staff Emeritus
With the correction given by Mark44, the result is:
$\displaystyle f(x)=\frac{\sqrt{x}}{\sqrt{x+2}+\sqrt{2}}\ .$​
And indeed, the numerator is √(x) as you have.

I gather from your Original Post, that you need to come up with a δ, ε -proof.

So, given an arbitrary ε > 0, you need to show that there is a δ ( which usually depends upon ε ) such that for any x, in the domain of f, that satisfies 0 < |x - 0| < δ, it's true that |f(x) - L|< ε. In this case L = 0.

12. Feb 20, 2012

### vela

Staff Emeritus
Oops, forgot about the $\sqrt{x}$ on the bottom. Just ignore me.