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adam32885
lim as t approaches 1 ((t^4-1)/(t^3-1)
Do you know L'Hopital's rule yet?
Ok Well you can proceed this way: You want to try and get rid of as many powers as you can. Why not try dividing the numerator and denominator by t^3?
I know i need to factor the top and bottom but i am drawing a blank on factoring
oh duh so it would t-1?
oh wait that doesn't help sorry. I'm so used of thinking of limits going to infinity
Ok Yes I suppose factoring is another way
that is where we are currently at is trying to factor them out
Ok t^4-1 is easy to factor. If you think of letting t^2 = x, then you get (x^2-1) which is easily factorable.
t^3-1 is a bit trickier. Do you remember that one way to this is to try finding all factors of your constant term (here it's -1) and plugging it into t^3-1 and see which ones give you zero. This number (say "a") will then be a factor of the form (t-a). Then, you can use long polynomial division to find out the other factor by dividing t^3-1 with (t-a)
Let me know if you are still stuck after :)