What are the discontinuities of #f(x) = (x^4 - 1)/(x-1)#?

1 Answer
MeneerNask
May 26, 2015

The only real discontinuity is when the denominator #x-1=0#
But I think you call that a hole. The value of #y#, either approached from above or below #x=1# give the same limit.

You can rewrite the function:

#=((x^2+1)(x^2-1))/(x-1)=((x^2+1)(x+1)(x-1))/(x-1)#

and cancel out the #(x-1)#'s ( only if #x!=1#)

#=(x^2+1)(x+1)# which has no further discontinuities
graph{(x^4-1)/(x-1) [-10, 10, -5, 5]}

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