# Different Way to Solve a Limit of an Indeterminate Equation

1. Aug 14, 2008

### chislam

$$lim_{x\rightarrow-2}\frac{x^3 + 8}{x+2}$$

Ok, I have solved this using synthetic division in which the limit was 12, however I was wondering if there was a way to solve this without using synthetic division (seeing as I hate the process of it).

I've solved plenty of other indeterminate equations by factoring out a polynomial and then being able to cancel out so that I can then plug in x, but I wasn't able to do so with this one.

I'm just curious if there are any other ways to solve this specific limit.

Thanks

2. Aug 14, 2008

### rootX

L'hospital rule (applies only when it is in forms like 0/0, inf/inf)

differentiate both num and den, and then sub in your -2

3. Aug 14, 2008

### chislam

Yeah I started reading up on that rule, but didn't understand it fully until your reply. So the derivative of the numerator is 3x^2 and the denominator is 1 so then just sub in the -2 and you get 12. Got it, thanks a lot for clearing up that rule for me.

4. Aug 14, 2008

### dynamicsolo

Not that this is much different than using synthetic division, but the numerator is a sum of two cubes, which factors according to

$$a^3 + b^3 = (a + b) \cdot (a^2 - ab + b^2)$$ . So, in fact, this numerator can be factored and the (x + 2) term can be cancelled.

Last edited: Aug 14, 2008