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

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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#

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