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How do you evaluate the limit #lim (e^t-1)/sint# as #t->0#?

Answer 1

Emilia Aguilar

Currently we're in indeterminate form, #0/0#, so we can use l'Hopitals rule to solve the problem.

#lim_(x-> c) (f(x))/(g(x)) = lim_(x-> c) (f'(x))/(g'(x))#

#lim_(t-> 0) (e^t - 1)/sint = lim_(t -> 0) (e^t)/cost#

#=e^0/cos(0)#

#= 1/1#

#= 1#

Hopefully this helps!

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Answer 2

Scarlett Allison

To evaluate the limit lim (e^t-1)/sint as t->0, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get lim (e^t)/cost as t->0. Plugging in t=0, we have lim (e^0)/cos0 = 1/1 = 1. Therefore, the limit is equal to 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you evaluate the limit #lim (e^t-1)/sint# as #t-&gt;0#?

How do you evaluate the limit #lim (e^t-1)/sint# as #t->0#?