`lim_(x->0) (e^x - 1)/(tan(x))` Evaluate the limit.

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Chapter 4, Review - Problem 7 - 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 x in expression under the limit, such that:

`lim_(x->0) (e^(x) - 1)/(tan x) = (e^0 - 1)/(tan 0) = (1-1)/0 = 0/0`

Hence, since the result is indeterminate `0/0, ` you may use l'Hospital's theorem, such that:

`lim_(x->0)(e^(x) - 1)/(tan x) = lim_(x->0) ((e^(x) - 1)')/((tan x)')`

`lim_(x->0) ((e^(x) - 1)')/((tan x)') = lim_(x->0) (e^x)/(1 + tan^2 x)`

Replacing 0 for x yields:

`lim_(x->0) (e^x)/(1 + tan^2 x) = (e^0)/(1 + tan^2 0) = 1/(1+0) = 1`

Hence, evaluating the given limit yields `lim_(x->0) (e^(x) - 1)/(tan x) = 1.`

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