How do you evaluate the limit #(x^4-16)/(x-2)# as x approaches #2#?
1 Answer
Aug 29, 2016
Explanation:
#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
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#
