Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Do you know L'Hopital's rule yet?
no not 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?
Not the answer you are looking for? Search for more explanations.
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)