Calculus: Early Transcendentals, Chapter 4, 4.4, Section 4.4, Problem 65

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).