How do you evaluate the limit $(x^4-16)/(x-2)$ as x approaches $2$ ?

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Answer 1 · 2016-08-29T21:54:26

$lim_(x->2) (x^4-16)/(x-2) = 32$

$x^4-16$

$=(x^2)^2-4^2$

$=(x^2-4)(x^2+4)$

$=(x^2-2^2)(x^2+4)$

$=(x-2)(x+2)(x^2+4)$

So:

$(x^4-16)/(x-2) = (color(red)(cancel(color(black)((x-2))))(x+2)(x^2+4))/color(red)(cancel(color(black)((x-2)))) = (x+2)(x^2+4)$

with exclusion $x != 2$

Hence:

$lim_(x->2) (x^4-16)/(x-2)$

$=lim_(x->2) ((x+2)(x^2+4))=((2)+2)((2)^2+4)=4*8 = 32$

How do you evaluate the limit (x^4-16)/(x-2)(x^4-16)/(x-2) as x approaches 22?