Asked by Anonymous
Use l'Hospital's Rule to evaluate ((te^7t)/7)-((e^7t)/49)?
Answers
Answered by
Reiny
L'Hopital's Rule, not L'Hospital's, is used in the evaluation of limits.
Is this a limit question?
Here is a nice video by Sal of Khan Academy, with an introduction to it.
www.khanacademy.org/math/ap-calculus-ab/ab-diff-contextual-applications-new/ab-4-7/v/introduction-to-l-hopital-s-rule
Is this a limit question?
Here is a nice video by Sal of Khan Academy, with an introduction to it.
www.khanacademy.org/math/ap-calculus-ab/ab-diff-contextual-applications-new/ab-4-7/v/introduction-to-l-hopital-s-rule
Answered by
oobleck
((te^7t)/7)-((e^7t)/49) = e^(7t) (7t-1)/49
What is the limit on t?
t->0 : no problem
t->∞ : ∞
t->-∞ : 0
t -> 1/7 : 0
you need something like 0/0 or 0*∞ of ∞/∞ for l'Hospital's Rule
What is the limit on t?
t->0 : no problem
t->∞ : ∞
t->-∞ : 0
t -> 1/7 : 0
you need something like 0/0 or 0*∞ of ∞/∞ for l'Hospital's Rule
Answered by
Anonymous
t is the limit.Is there other way to solve this?
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