# `lim_(x->0^+) (cos(x))^(1/x^2)` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t...

`lim_(x->0^+) (cos(x))^(1/x^2)` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

### Textbook Question

Chapter 4, 4.4 - Problem 65 - 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:

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

You may use the special limit `lim_(x->0^+) (1 + x)^(1/x) = e` , instead of l'Hospital's rule, such that:

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

Evaluate the limit of exponent, such that:

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

You may use l'Hospital's rule for indetermination `0/0` , such that:

`lim_(x->0^+)((cos x - 1)')/((x^2)') = lim_(x->0^+)(-sin x)/(2x) = -(sin0)/0 = 0/0`

You may use again l'Hospital's rule for indetermination `0/0` , such that:

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

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

Hence, evaluating the limit, using special limit and l'Hospital's rule, yields `lim_(x->0^+) (cos x)^(1/(x^2)) = 1/(sqrt e).`