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How calculate limit F(cos x )-F(1)/x^2, x->0? f(x)=e^-x^2

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uoor | Student, College Freshman | eNoter

Posted May 19, 2013 at 1:39 PM via web

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How calculate limit F(cos x )-F(1)/x^2, x->0? f(x)=e^-x^2

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted May 19, 2013 at 2:13 PM (Answer #1)

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You need to evaluate the following limit such that:

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

Replacing 0 for x yields:

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

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

The indetermination 0/0 requests for you to use l'Hospital's theorem, such that:

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

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

You may split the limit such that:

`lim_(x->0) (f(cos x)*(-sin x))/(2x)= lim_(x->0) (f(cos x))/2*lim_(x->0)(-sin x)/x`

Since the remarcable limit `lim_(x->0)(sin x)/x = 1` yields:

`lim_(x->0) (f(cos x)*(-sin x))/(2x) = -1/2*f(cos 0)`

`lim_(x->0) (f(cos x)*(-sin x))/(2x) = -1/2f(1)`

Replacing 1 for x in equation of the function yields:

`lim_(x->0) (f(cos x)*(-sin x))/(2x) = -1/2*(e^(-1^2))`

`lim_(x->0) (f(cos x)*(-sin x))/(2x)= -1/(2e)`

Hence, evaluating the given limit using l'Hospital's theorem, yields `lim_(x->0) (F(cos x) - F(1))/x^2 = -1/(2e).`

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