# Solving lim(x->0)((cosx)^(-4/(x^2))) without L'Hopital's Rulelim(x->0)((cosx)^(-4/(x^2))) I would like to know how I can solve it without using De L'Hospital's rule....

Solving lim(x->0)((cosx)^(-4/(x^2))) without L'Hopital's Rule

lim(x->0)((cosx)^(-4/(x^2)))

I would like to know how I can solve it without using De L'Hospital's rule.

http://img803.imageshack.us/img803/7954/wolframalphalimxgt0cosx.png

Thanks!

### 1 Answer | Add Yours

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.` **