# Limit of a fraction

1. Jan 26, 2008

### fermio

1. The problem statement, all variables and given/known data
$$\lim_{x\to 6}\frac{\sqrt{x+3}-3}{x-6}$$

2. Relevant equations

3. The attempt at a solution
$$\lim_{x\to 6}\frac{\sqrt{x+3}-3}{x-6}=\lim_{x\to 6}\frac{(x+3-9)(x+6)}{(x^2-36)(\sqrt{x+3}+3)}$$

2. Jan 26, 2008

### Mathdope

Look more carefully at your attempt at a solution. You actually have it.

3. Jan 26, 2008

### arildno

Do not bother with expanding with (x+6) as well:
$$\frac{\sqrt{x+3}-3}{(x-6)}=\frac{x+3-9}{(x-6)(\sqrt{x+3}+3)}=\frac{(x-6)}{(x-6)(\sqrt{x+3}+3)}$$

4. Jan 26, 2008

### fermio

"Look more carefully at your attempt at a solution. You actually have it."
I don't see.
$$\lim_{x\to 6}\frac{\sqrt{x+3}-3}{x-6}=\lim_{x\to 6}\frac{(x+3-9)(x+6)}{(x^2-36)(\sqrt{x+3}+3)}=\lim_{x\to 6}\frac{x^2+6x-6x-36}{x^2\sqrt{x+3}+3x^2-36\sqrt{x+3}-108}$$

I get it.
$$\lim_{x\to 6}\frac{\sqrt{x+3}-3}{x-6}=\lim_{x\to 6}\frac{x+3-9}{(x-6)(\sqrt{x+3}+3)}=\lim_{x\to 6}\frac{1}{\sqrt{x+3}+3}=\frac{1}{6}$$

Last edited: Jan 26, 2008
5. Jan 26, 2008

### Rainbow Child

How can you write $x^2-36$?