Lim as X approaches 2 (rationalizing wrong)

[r6mikey]

Homework Statement

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

Homework Equations

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!

 

[NateTG]

Homework Statement

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

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!

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.

 

[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

 

[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 Xand 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