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kirbykirbyBest ResponseYou've already chosen the best response.1
Do you know L'Hopital's rule yet?
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
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?
 one year ago

adam32885Best ResponseYou've already chosen the best response.0
I know i need to factor the top and bottom but i am drawing a blank on factoring
 one year ago

adam32885Best ResponseYou've already chosen the best response.0
oh duh so it would t1?
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
oh wait that doesn't help sorry. I'm so used of thinking of limits going to infinity
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
Ok Yes I suppose factoring is another way
 one year ago

adam32885Best ResponseYou've already chosen the best response.0
that is where we are currently at is trying to factor them out
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
Ok t^41 is easy to factor. If you think of letting t^2 = x, then you get (x^21) which is easily factorable.
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
t^31 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^31 and see which ones give you zero. This number (say "a") will then be a factor of the form (ta). Then, you can use long polynomial division to find out the other factor by dividing t^31 with (ta)
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
Let me know if you are still stuck after :)
 one year ago
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