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

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Chapter 4, Review - Problem 10 - 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 `oo`   for x in expression under the limit, such that:

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

Hence, since the result is indeterminate `(oo)/(oo)` , 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 `oo` for x yields:

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

Hence, since the result is indeterminate `(oo)/(oo)` , 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 `oo` for x yields:

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

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

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