Question

l'hopital rule:

lim as t approaches 0

(e^2t-1)/(sin t)
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so when i plug in t as 0, i get 0/0 therefore i can apply l'hopital's rule. once i take the derivative, i get (e^2t)/(cos t). taking the limit to zero, i get 1/1 which equals 1, but the book says 2.

Answer 1

No there is a chain rule piece when you apply l'hopital's rule. You will get (2e^(2t))/(cos t). If you take the limit to 0, you get 2

Answer 2

so the derivative of e^x is e^x right? if its e^2x i chain it so it becomes 2e^x?

Answer 3

No, e^(2x) becomes 2e^(2x)

Answer 4

would you like me to explain why?

Answer 5

yes please

Answer 6

oh nevermind

Answer 7

i got it

Answer 8

thank you so much