# Limits

• Apr 3rd 2006, 06:37 PM
becky
Limits
Just need a couple of problems double checked for me.

lim (x^2+4x-5)/x^2-1
x->1

I am saying that this limit does not exist because of the 0 numberator.

In this one, I have forgotten the rules for square roots.

lim [(sq rt 2x-1) -3]/x-5
x->5-
• Apr 3rd 2006, 06:57 PM
ThePerfectHacker
Quote:

Originally Posted by becky
Just need a couple of problems double checked for me.

lim (x^2+4x-5)/x^2-1
x->1

You have,
$\displaystyle \lim_{x\to 1}\frac{x^2+4x-5}{x^2-1}$
Express as,
$\displaystyle \frac{(x-1)(x+5)}{(x-1)(x+1)}$
Which cancels,
$\displaystyle \lim_{x\to 1}\frac{x+5}{x+1}=\frac{6}{2}=3$
• Apr 3rd 2006, 07:03 PM
ThePerfectHacker
Quote:

Originally Posted by becky

lim [(sq rt 2x-1) -3]/x-5
x->5-

You have,
$\displaystyle \lim_{x\to 5}\frac{\sqrt{2x-1}-3}{x-5}$
Rationalize numerator,
$\displaystyle \frac{\sqrt{2x-1}-3}{x-5}\cdot \frac{\sqrt{2x-1}+3}{\sqrt{2x-1}+3}$
Thus,
$\displaystyle \frac{2x-10}{(x-5)(\sqrt{2x-1}+3)}$
Thus,
$\displaystyle \frac{2(x-5)}{(x-5)(\sqrt{2x-1}+3)}$
Thus,
$\displaystyle \lim_{x\to 5}\frac{2}{\sqrt{2x-1}+3}=\frac{2}{3+3}=1/3$
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Basic rule, is that if you have a fraction that does not lead to division by zero upon substitution then you can substitute.