Find the limit if it exists: (x^4-81)/(x-3) x-3

Question

Find the limit if it exists: (x^4-81)/(x-3) x->3

anonymous · Answer

$$x^4-3^4=(x^2+3^2)(x^2-3^2)=(x^2+3^2)(x-3)(x+3)$$
So, what you really have is 

$$\frac{(x^2+3^2)(x-3)(x+3)}{x-3}=(x^2+3^2)(x+3) \;\; x\rightarrow3 $$
$$(3^2+3^2)(3+3)=18*6=108$$
and Bingo.

anonymous · Answer

hahahahhahahahhahahaha, how did you get yours to look like that?

nottim · Answer

I'm Batman.

anonymous · Answer

You can factor an (x-3) out of the numerator, which cancels out the (x-3) in the denominator. Then just plug in three to find the limit. This evaluates to 108.

anonymous · Answer

i thought you said earlier if the numerator has a higher power than the denom, then the limit doesn't exist?

anonymous · Answer

That only applies when x -> infinity

anonymous · Answer

Yes, that only applies when x->infinity. ^^ 
Tricks of the trade.