Solving Tuff Limit Problem: Factoring Out Rational Function

[Alem2000]

Okay basically i tried to facter out the top part of the rational function but it just doesn't 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.