Question

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

Answer 1

`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`

Answer 2

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`