Question

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

Answer 1

\[x^4-3^4=(x^2+3^2)(x^2-3^2)=(x^2+3^2)(x-3)(x+3)\]
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.

Answer 2

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

Answer 3

I'm Batman.

Answer 4

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.

Answer 5

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

Answer 6

That only applies when x -> infinity

Answer 7

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