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kirbykirby
 2 years ago
Best ResponseYou've already chosen the best response.1Do you know L'Hopital's rule yet?

kirbykirby
 2 years ago
Best ResponseYou've already chosen the best response.1Ok 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?

adam32885
 2 years ago
Best ResponseYou've already chosen the best response.0I know i need to factor the top and bottom but i am drawing a blank on factoring

adam32885
 2 years ago
Best ResponseYou've already chosen the best response.0oh duh so it would t1?

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

kirbykirby
 2 years ago
Best ResponseYou've already chosen the best response.1Ok Yes I suppose factoring is another way

adam32885
 2 years ago
Best ResponseYou've already chosen the best response.0that is where we are currently at is trying to factor them out

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

kirbykirby
 2 years ago
Best ResponseYou've already chosen the best response.1t^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)

kirbykirby
 2 years ago
Best ResponseYou've already chosen the best response.1Let me know if you are still stuck after :)
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