Use l'Hospital's Rule to evaluate ((te^7t)/7)-((e^7t)/49)?

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

((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

t is the limit.Is there other way to solve this?

To evaluate the expression ((te^7t)/7)-((e^7t)/49) using l'Hospital's Rule, we need to determine the limit as t approaches 0.

First, let's differentiate the numerator and denominator separately:

The derivative of te^7t with respect to t is given by:

d(te^7t)/dt = e^7t + 7te^7t

Similarly, the derivative of e^7t with respect to t is:

d(e^7t)/dt = 7e^7t

Now, let's substitute these derivatives into the expression:

lim t→0 ((te^7t)/7)-((e^7t)/49)
= lim t→0 ((e^7t + 7te^7t)/7) - ((7e^7t)/49)

Next, let's find the limit of the expression as t approaches 0. We can evaluate it by directly substituting t = 0 into the expression:

((e^7(0) + 7(0)e^7(0))/7) - ((7e^7(0))/49)

Simplifying this expression gives:

(1/7) - (7/49)
= 1/7 - 1/7
= 0

Therefore, using l'Hospital's Rule, we find that the value of ((te^7t)/7)-((e^7t)/49) as t approaches 0 is 0.