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

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

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

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

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