# Math Help - Evaluating limits.

1. ## Evaluating limits.

Evaluate the limit, if it exists. (If it does not exist, enter NONE).

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.

2. I think the next step would look somethin like this..

(x-2)(x+3)
--------------
x-2

...but im not exactly positive.

3. That's correct. Continue on.

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

5. Remember that when x= a makes polynomial P(x) zero- that is if P(a)= 0, then x-a 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, $2^2+ 2- 6= 0$, you know that you can factor out x-2. Once you know that, it should be clear that $x^2+ x- 6= (x- 2)(x+ 3)$.

Now you can think $\frac{x^2+ x- 6}{x- 2}= \frac{(x-2)(x+3)}{x-2}= x+ 3$ as long as $x\ne 2$.

Notice that "as long as $x\ne 2$". Of course, $\frac{x^2+x-6}{x-2}$ is "undefined" at x= 2 while x+3 is equal to 5 at x= 2 so we cannot just say that $\frac{x^2+ x- 6}{x- 2}= x+3$. Fortunately there is an (often overlooked) theorem about limits that says "If f(x)= g(x) for all x except x=a, then $\lim_{x\to a} f(x)= \lim_{x\to a} g(x)$".

Knowing that $\frac{x^2+x-6}{x-2}= x+ 3$ for all x except x= 2 is enough to tell us they have the same limit at x= 2 and, of course, $\lim_{x\to 2} x+ 3$ is easy.

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