`lim_(t->0) (e^(2t) - 1)/(sin(t))` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t...

`lim_(t->0) (e^(2t) - 1)/(sin(t))` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

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Textbook Question

Chapter 4, 4.4 - Problem 13 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the limit, hence, you need to replace 0 for t in the limit, such that:

`lim_(t->0) (e^(2t) - 1)/(sin t) = (e^0 - 1)/(sin 0) = (1-1)/0 = 0/0`

Since the limit is indeterminate `0/0` , you may use l'Hospital's rule:

`lim_(t->0) (e^(2t) - 1)/(sin t) = lim_(t->0) ((e^(2t) - 1)')/((sin t)') `

`lim_(t->0) ((e^(2t) - 1)')/((sin t)')= lim_(t->0) (2e^(2t))/(cos t)`

You need to replace 0 for t, such that:

`lim_(t->0) (2e^(2t))/(cos t) = (2e^(0))/(cos 0) = (2*1)/1 = 2`

Hence, evaluating the given limit, using l'Hospital's rule, yields `lim_(t->0) (e^(2t) - 1)/(sin t) = 2.`

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