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

Asked on by enotes

### Textbook Question

Chapter 4, Review - Problem 8 - Calculus: Early Transcendentals (7th Edition, James Stewart).
See all solutions for this textbook.

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

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.`

We’ve answered 319,620 questions. We can answer yours, too.