lim
x->c

hard to see that part
looking for a walkthrough/explanation more than the answer


This post was edited by Illumini on Jan 29 2014 09:17pm

The limit exists anywhere where you don't have a divide by zero. You find the factors of the top and those factors tell you what c cannot be.

In this case, c cannot be 2, -2, or -3, so the limit exists everywhere else.

Not sure if you have to worry about infinite limits but if you do there is an additional caveat.

This post was edited by Dontrunaway on Jan 29 2014 09:27pm

not 100% true, you'll have to look at graph/limits from left/right for some equ

This post was edited by saber_x3 on Jan 29 2014 10:22pm

Have to do it algebraically without the use of L'hospitals rule as well

Given the factorization above, if c is different from 2, -2 and -3, your quotient will have 2 different limits when x -> c by the left or by the right.
The numerator will tend to a given non-zero real number, whereas the denominator will tend to zero with negative values or with positive values. Hence limit = - infinity or + infinity.
Hence, the limit as x-> c doesn't exist.

On the contrary, if c = 2, c = - 2 or c = - 3, then after a cancellation it's easy to see that your limit exists.

It would be c = 3, to cancel wouldn't it