How do you find the limit of (t^2-4)/(t^3-8) as t approaches 2?

2 Answers
Krzysztof S.
Nov 17, 2016

lim f(t) _(t=2)=1/3

Explanation:

f(t) = (t^2-4)/(t^3-8) for t=2 gives 0/0

is possible to use L'Hôpital's rule

lim f(t) _(t=2)=lim(2t)/(3t^2)= lim 2/(3t)=1/3

Steve M
Nov 17, 2016

An algebraic solution:

lim_(t rarr 2) (t^2-4)/(t^3-8) = lim_(t rarr 2) (t^2-2^2)/(t^3-2^3)
lim_(t rarr 2) (t^2-4)/(t^3-8) = lim_(t rarr 2) ((t+2)(t-2))/((t-2)(t^2+2t+4))
lim_(t rarr 2) (t^2-4)/(t^3-8) = lim_(t rarr 2) ((t+2))/((t^2+2t+4))
lim_(t rarr 2) (t^2-4)/(t^3-8) = (2+2)/(4+4+4)
lim_(t rarr 2) (t^2-4)/(t^3-8) = (4)/(12)
lim_(t rarr 2) (t^2-4)/(t^3-8) = 1/3

Related questions
See all questions in Determining Limits Algebraically
Impact of this question
8603 views around the world
You can reuse this answer
Creative Commons License