`lim_(x->0) tan(4x)/(x + sin(2x))` Evaluate the limit.

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Chapter 4, Review - Problem 8 - 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) (tan 4x)/(x + sin 2x) = (tan 0)/(0 + sin 0) = 0/0`

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

`lim_(x->0) (tan 4x)/(x + sin 2x)= lim_(x->0) ((tan 4x)')/((x + sin 2x)')`

`lim_(x->0) ((tan 4x)')/((x + sin 2x)') = lim_(x->0) (4/(cos^2 4x))/(1 + 2cos 2x)`

Replacing 0 for x yields:

`lim_(x->0) (4/(cos^2 4x))/(1 + 2cos 2x) = (4/(cos^2 0))/(1 + 2cos 0)`

`lim_(x->0) (4/(cos^2 4x))/(1 + 2cos 2x) =(4/1)/(1 + 2) = 4/3`

Hence, evaluating the given limit yields `lim_(x->0) (tan 4x)/(x + sin 2x) = 4/3.`

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