Tuff limit problem

[Alem2000]

Okay basically i tried to facter out the top part of the rational function but it just doesnt seem corect.

\lim{x\rightarrow2}

\frac{x^4-16}{x-2}

\lim{x\rightarrow2}

\frac{(x^2+4)(x^2-4}{X-2}

\lim{x\rightarrow2} \frac{(x+2)(x-2)(x^2-4)}{x-2}

\lim{x\rightarrow2} (x+2)(x^2-4)=0
IS that correct, is it okay to factor out one and cancel?

[Gokul43201]

Surely you mean

\lim_{x\rightarrow2}~~\frac{(x^2-4)(x^2+4)}{x-2}

= \lim_{x\rightarrow2}~~ \frac{(x+2)(x-2)(x^2+4)}{x-2}

=\lim_{x\rightarrow2}~~ (x+2)(x^2+4)=32

It is okay to factor out and cancel.

[Allah]

yep, gokul is correct

[Alem2000]

Hmm, sloppy hand writing on my paper seems to be the cause of confusion.
after simplyfing (x^2-4) I thought I had simplified (x^2+4) ...Even though this is a limit problem my gratatude is limitless... :rofl:
yeah i konw its corny but i had to say it...thanks a million though Gokul43201.

[HallsofIvy]

The reason it is "correct to cancel" is that if f(x) and g(x) have the same value every where except at x= a, then \lim_{x\rightarrow a}f(x)= \lim_{x\rightarrow a}g(x).

In this problem, as long as x is not 2, \frac{x^4-16}{x-2}= (x^2+4)(x^2+2) so the limits at 2 are the same.