# Simple Limit Problem

1. Jun 21, 2006

### es

I tutor high school students in Calc and the other day I came across this problem.

Limit of (1-Sqrt(x-2))/(x-3) as x->3

I tried coaching the student on how to simplify the expression and in the end I just showed him this substitution.

Let u=Sqrt(x-2)

Then

(1-Sqrt(x-2))/(x-3) = (1-u)/(u^2-1)

And the Limit becomes

Limit of (1-u)/(u^2-1)=-1/(u+1) as u->1 which is -1/2

He looked at me like I had just done some black magic. I explained substitution to him and why it worked, showed him a couple of other simple examples, and confirmed the answer numerically (like they do in basic calc books when the limit concept is first presented). I still don't think he is 100% convinced because they had not covered this in his class yet which leads to my question.

Can the original problem be solved without substitution?

2. Jun 21, 2006

### arildno

Well, you might do it this way:
$$\frac{(1-\sqrt{x-2})}{x-3}=\frac{(1-\sqrt{x-2})}{x-3}*1=\frac{(1-\sqrt{x-2})}{x-3}*\frac{(1+\sqrt{x-2})}{(1+\sqrt{x-2})}=\frac{3-x}{(1+\sqrt{x-2})*(x-3)}=-\frac{1}{(1+\sqrt{x-2})}$$
And so on..

3. Jun 21, 2006

### arunbg

Multiplying by the conjugate would do the trick, but personally, I feel the easiest way is L'Hospital's rule .

4. Jun 22, 2006

### es

Yipe. For some reason I though 1-x would be the numerator. Boy do I feel sheepish. :) Thanks for the help!