You need to take logarithm of the limit such that:

`ln lim_(x->0) ((cosx)^(-4/(x^2))) => lim_(x->0) ln ((cosx)^(-4/(x^2)))`

You need to use the logarithmic identity such that:

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

You may use l'Hospital's theorem such that:

`lim_(x->0) ((-4 ln (cos x))')/((x^2)') = lim_(x->0) (-4(-sin x)/cos x)/(2x)`

You need to substitute `tan x` for `sin x/cos x` such that:

`lim_(x->0) (-4(-sin x)/cos x)/(2x) = (4/2) lim_(x->0) tan x/x`

You need to use the remarcable limit such that:

`lim_(x->0) tan x/x = 1`

`lim_(x->0) -4/(x^2) ln (cos x) = (4/2)*1`

`lim_(x->0) -4/(x^2) ln (cos x) = 2`

You need to remember that `ln lim_(x->0) ((cosx)^(-4/(x^2))) = 2 =>lim_(x->0) ((cosx)^(-4/(x^2))) = e^2` .

**Hence, evaluating the given limit using logarithmic identities yields `lim_(x->0) ((cosx)^(-4/(x^2))) = e^2.` **

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.