# `lim_(x->0) (e^(4x) - 1- 4x)/(x^2)` Evaluate the limit.

### Textbook Question

Chapter 4, Review - Problem 9 - 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) (e^(4x) - 1 - 4x)/(x^2) = (e^0 - 1 - 0)/(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^(4x) - 1 - 4x)/(x^2) = lim_(x->0) ((e^(4x) - 1 - 4x)')/((x^2)')`

`lim_(x->0) ((e^(4x) - 1 - 4x)')/((x^2)') = lim_(x->0) (4e^(4x) - 4)/(2x)`

Replacing 0 for x yields:

`lim_(x->0) (4e^(4x) - 4)/(2x) = (4e^0 - 4)/(2*0) = (4-4)/0 = 0/0`

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

`lim_(x->0) (4e^(4x) - 4)/(2x) = lim_(x->0) ((4e^(4x) - 4)')/((2x)')`

`lim_(x->0) ((4e^(4x) - 4)')/((2x)') = lim_(x->0) (16e^(4x))/2`

Replacing 0 for x yields:

`lim_(x->0) (16e^(4x))/2 = (16e^0))/2 = 16/2 = 8`

Hence, evaluating the given limit yields `lim_(x->0) (e^(4x) - 1 - 4x)/(x^2) = 8.`