Asked by victor.m

Evaluate limit squareroot of (x^4 - 16)/(x^3 - 8) as x approaches 2

Answers

Answered by drwls
x^4 -16 = (x^2 +4)(x+2)(x-2)
x^3 -8 = (x-2)(x^2 +2x + 4)
The x-2 terms cancel
(x^4-16)/(x^3-8)
= (x^2+4)(x+2)/(x^2+2x+4)
-> 8*4/12 = 8/3 as x -> 2

If you are familiar with calculus, you could also take the ratio of derivatives of numerator and denominator, and get the same answer.
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