Asked by Sarah
The limit as x approaches 1 (x^3-1)/(x^2-1) is
I got 3/2
I got 3/2
Answers
Answered by
MathMate
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.
(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.