# Limits involving square roots

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• August 29th 2012, 09:18 AM
nikdsm2012
Limits involving square roots
Hello. I've never been great at math (or spent much time studying), but, since my new major requires quite a few math classes, I have decided to put in as much time as I can to develop my skills. I need help finding a limit for a problem involving square roots. The problem is probably very simple, but I am not sure exactly how to work with the square roots/squares in the problem. I have tried as many approaches as I can think of, and I still feel like I'm getting nowhere.

Could anyone walk me through the steps in solving this? That would be EXTREMELY helpful. The problem:

lim t->2

sqrt( (t+4)(t-2)^4 )
-------------------------
(3t - 6)^2

Sorry, I have no idea how to format that properly. Thanks for any help in advance!
• August 29th 2012, 09:30 AM
Plato
Re: Limits involving square roots
Quote:

Originally Posted by nikdsm2012
Hello. I've never been great at math (or spent much time studying), but, since my new major requires quite a few math classes, I have decided to put in as much time as I can to develop my skills. The problem:

lim t->2

sqrt( (t+4)(t-2)^4 )
-------------------------
(3t - 6)^2N

No walk through is available here. We don't do that.

But $\(3t-6)^2= 9(t-2)^2$ so $\frac{\sqrt{(t+4)(t-2)^4}}{(3t-6)^2}=\frac{\sqrt{(t+4)}(t-2)^2}{9(t-2)^2}=\frac{\sqrt{(t+4)}}{9}$
• August 29th 2012, 09:39 AM
ebaines
Re: Limits involving square roots
If this is what you mean:

$\lim _{t \to 2} \frac {\sqrt{(t+4)(t-2)^4}}{(3t-6)^2}$

You can start by pulling the (t-2)^4 term out of the square root, then simplify:

$\frac {\sqrt{(t+4)(t-2)^4}}{(3t-6)^2} = \frac {\sqrt{t+4}(t-2)^2}{3^2(t-2)^2} = \frac {\sqrt{t+4}} 9$

$\lim _{t \to 2} \frac {\sqrt{t+4}} 9 = \frac {\sqrt 6 } 9$