
Evaluating limits.
Evaluate the limit, if it exists. (If it does not exist, enter NONE).
http://www.webassign.net/cgibin/sym...29%2F%28x2%29
I know according to rule 5 of the law of limits, this cant work by substition because the denominator would equal 0. I know i need to start this off by factoring but im not sure how to start this process.
Thanks.

I think the next step would look somethin like this..
(x2)(x+3)

x2
...but im not exactly positive.

That's correct. Continue on.

I got it, i made a mistake swapping a + sign for a .... Thank you!

Remember that when x= a makes polynomial P(x) zero that is if P(a)= 0, then xa must be a factor of P(x). Here, you saw correctly that the denominator is 0 when x= 2. That should make you look immediately at the numerator!
If the numerator were NOT 0 at x= 2, that would tell you that the limit does not exist and you can stop.
If the numerator is 0 at x= 2, which is the case here, $\displaystyle 2^2+ 2 6= 0$, you know that you can factor out x2. Once you know that, it should be clear that $\displaystyle x^2+ x 6= (x 2)(x+ 3)$.
Now you can think $\displaystyle \frac{x^2+ x 6}{x 2}= \frac{(x2)(x+3)}{x2}= x+ 3$ as long as $\displaystyle x\ne 2$.
Notice that "as long as $\displaystyle x\ne 2$". Of course, $\displaystyle \frac{x^2+x6}{x2}$ is "undefined" at x= 2 while x+3 is equal to 5 at x= 2 so we cannot just say that $\displaystyle \frac{x^2+ x 6}{x 2}= x+3$. Fortunately there is an (often overlooked) theorem about limits that says "If f(x)= g(x) for all x except x=a, then $\displaystyle \lim_{x\to a} f(x)= \lim_{x\to a} g(x)$".
Knowing that $\displaystyle \frac{x^2+x6}{x2}= x+ 3$ for all x except x= 2 is enough to tell us they have the same limit at x= 2 and, of course, $\displaystyle \lim_{x\to 2} x+ 3$ is easy.

great tip, i wasnt aware of that trick to determine if there is a limit! Awsome, i love this place.