calculate lim x -> 0 of (cos(x))^(1/x^2)?

Question

whats the limit of cosx^1/x^2 as x goes to 0?

matlab says exp(-1/2)
but obviously i'll need a prove and explanation.

Answer 1

y = (cosx)^(1/x²)



ln(y) =Lim 1/x² ln(cosx)



use L'hospital rule:

ln(y) = Lim d/dx ln(cosx) / d/dx (x²)



ln(y) = Lim (-sinx/cosx) / (2x)



ln(y) = Lim (-tanx) / 2x



use L'hosptial again

ln(y) = Lim (-sec²x)/2



ln(y) = Lim (-sec²0)/2



ln(y) = Lim -1/2



y = Lim e^(-1/2)

Answer 2

Let f(x) = (cos(x))^(1/x^2)



We are looking for lim x -> 0 f(x)



ln(lim x -> 0 f(x)) = lim x -> 0 ln f(x)

= lim x -> 0 ln (cos(x))^(1/x^2)

= lim x -> 0 (1/x^2) * ln (cos x) //exponential rule for logarithms

= lim x -> 0 (ln(cos x))/(x^2)

= lim x -> 0 ((1 / cos x) * (-sin x)) / (2x) //L'Hopital's rule, as the previous form is a 0/0 form

= lim x -> 0 (-tan x) / 2x //tan = sin / cos

= lim x -> 0 (-sec^2 x) / 2 //L'Hopital's rule again, 0/0

= -1/2 //you can substitute now, as it is no longer 0/0



Thus, lim x -> 0 f(x) = exp(ln(lim x -> 0 f(x))) = exp(-1/2)

Answer 3

lim x-> 0 sine(x)/x = a million so sine(x)/x < a million => a million/x < a million/sine(x) => a million/x - a million/sine(x) < 0 lim x -> 0 ((a million/x)-(a million/sinx)) = lim x-> 0 (a million/x) - lim x-> 0 a million/sine(x) lim x-> 0 (a million/x) = ? lim x-> 0 a million/sine(x) = ? => lim x-> 0 (a million/x) - lim x-> 0 a million/sine(x) = 0^(-) => lim x -> 0 ((a million/x)-(a million/sinx)) = 0^(-) 0^(-) is decrease than 0 and procedures to 0