L'Hospital's rule

[KLscilevothma]

Q1) lim (cosx)(1/x^2)
x-->0

I tried to find

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

but got stuck after differentiate it several times.

Thanks in advance

[bogdan]

L'Hospital is quite complicated to be applied here...

(cosx)^(1/x^2)=[1-(sinx)^2]^(1/(2*(x^2)))=

={[1-(sinx)^2]^[-1/((sinx)^2)]}^(-((sinx)^2)/(2*(x^2)))

lim {[1-(sinx)^2]^[-1/((sinx)^2)]}^(-((sinx)^2)/(2*(x^2)))=
x->0

=lim {[1-(sinx)^2]^[-1/((sinx)^2)]}^lim(-((sinx)^2)/(2*(x^2)))=
x->0____________________________x->0
=e^(-1/2)=1/sqrt(e);

I used sinx/x -> 1 as x->0 and

lim (1+Xn)^(1/Xn)=e, where
x->0

lim Xn = 0, Xn >0 or Xn <0...
n->infinity

I must remember you L'Hospital's rule :

lim f(x)/g(x)=lim f'(x)/g'(x),
x->a_______x->a

and
lim f(x)=0 or infinity
x->a
lim g(x)=0 or infinity
x->a
...

[KLscilevothma]

Thank you bogdan!

I just got the answer before viewing your thread. I made a very silly mistake when doing this question at the very beginning.

Here is my approach.
lim (cosx)(1/x2)
x-->0

Let y=(cosx)(1/x2)
take natural log on both sides
ln y = (1/x2) ln cos x

Lim (1/x2) ln cos x
x-->0

=Lim ln(cosx)/x2 (which is an indeterminate form of 0/0)
x-->0

=Lim -sinx/(2xcosx)
x-->0

=Lim -(sinx)/x * 1/(2cosx)
x-->0

=-1/2

Therefore
lim (cosx)(1/x^2) = e-1/2
x-->0

[bogdan]

Very nice...and simple...