Any chance you can help me on this one mate? lim t -> 4 (4-t)/(5-[sqrt t^2 + 9]) i subbed in x = 4 and solved and got 0. right?
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Both the numerator and denominator goes to 0 as t goes to 4, so you can't just plug in . L'Hospitals rule works here though, so I suggest you use it.
When t=4 you don't get 0 you get 0/0 which isn't defined.
Originally Posted by Hooperoo Any chance you can help me on this one mate? lim t -> 4 (4-t)/(5-[sqrt t^2 + 9]) i subbed in x = 4 and solved and got 0. right? You can also do the following: Proceed ...
Last edited by Siron; Mar 24th 2012 at 06:12 AM.
Originally Posted by Siron You can also do the following: Proceed ... Why has 4-t changed to t-4?
Originally Posted by biffboy Why has 4-t changed to t-4? It was a typo, thanks for noticing that!
Originally Posted by Siron You can also do the following: Proceed ... Essentially, cant divide by 0, so limit is not defined?
You can factore
Originally Posted by Siron You can factore so im getting lim t -> 4 (4-t)(5-sqrt t^2+9)/(4-t)(4+t) do i then just cancel the (4-t) and solve like that?
Indeed.
$x_0=4$ is is called a removable discontinuity point.
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