The limit as x approaches 1 (x^3-1)/(x^2-1) is

I got 3/2

1 answer

First step, substitute x=1 to get
(x^3-1)/(x^2-1)=0/0 need to do something.

Then since both numerator and denominator are factorizable readily, we have
Limit (x->1) (x^3-1)/(x^2-1)
=Limit(x->1) (x-1)(x^2+x+1)/[(x-1)(x+1)]
=Limit(x->1) (x^2+x+1)/(x+1) since x≠1
=limit (x->1) (1+1+1)/(1+1) = 3/2
Yes, 3/2 is correct.

You can also use d'Hôpital's rule with the same result.
Similar Questions
  1. The limit as x approaches infinity. (1)/(5^x)The limit as x approaches 1. (1-x^3)/(2-sqrt(x^2-3)) Show your work thanks in
    1. answers icon 3 answers
  2. The table below gives selected values of a twice differentiable function f(x)x|. -7. -6. -4. -2. f(x)|. 0. -1. -2. 0 f'(x)|. 3.
    1. answers icon 1 answer
  3. Determine the behavior of limitsA. Limit as x approaches 1 of: (log x)/((x-1)^2) B. Limit as x approaches infinity of:
    1. answers icon 1 answer
    1. answers icon 0 answers
more similar questions