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-
You have,Originally Posted by becky
$\displaystyle \lim_{x\to 5}\frac{\sqrt{2x-1}-3}{x-5}$
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.