Lim as X approaches 2 (rationalizing wrong)

1. Sep 4, 2007

r6mikey

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

The limit as x approaches 2 for (X-2)/(sqrt7+x)-(x+1)

2. Relevant equations

3. The attempt at a solution
I know i have to rationalize the denominator but it seems like i'm doing something very wrong with my distrubution...please help!!!

2. Sep 4, 2007

NateTG

are X and x supposed to be distinct?

When writing things like this out, it's worth being a bit clearer, since what you've written could be:
$$\frac{X-2}{\sqrt{7}+x}-(x+1)$$
or
$$\frac{X-2}{(\sqrt{7}+x)-(x+1)}$$

Regardless, I don't see why you would need to rationalize the denominator.

3. Sep 4, 2007

r6mikey

lim as x approaches 2 for $$X-2/\sqrt{7+X}-(x+1)$$

this was the problem..I solved it to be -6/5......I just have a question...i have another similar problem, which also becomes in the indeterminate form.

lim as t approaches 3 for $$1-t+\sqrt{1+t}/t-3$$

where do i find more information on how to distribute here? I know i have to rationalize I am just lost in how distribution works with a problem with no parentheses and one with parentheses?

I have 4 different books here, 2 algebra, 2 calculus....and not sure what or where to review this

4. Sep 4, 2007

NateTG

In latex the construct for fractions is:
\frac{$numerator}{$denominator}
(You can click on the graphical version to see the code:
$$\frac{1}{4}$$
It will make things a bit more legible.

You seem to be using $X$and $x$ as if they were the same - they're not.

To rationalize:
$$\frac{x-2}{\sqrt{7+x}-(x+1)}$$
Multiply by:
$$\frac{\sqrt{7+x}+(x+1)}{\sqrt{7+x}+(x+1)}$$

Generally, if you have:
$$\sqrt{a} + b$$
you'll want to multiply by
$$\sqrt{a} - b$$
since this creates a difference of two squares:
$$(\sqrt{a} + b)\times(\sqrt{a} - b )=a-b^2$$