- #1

- 314

- 0

^{(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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter KLscilevothma
- Start date

- #1

- 314

- 0

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

- #2

- 191

- 0

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

...

(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

...

Last edited:

- #3

- 314

- 0

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/x

x-->0

Let y=(cosx)(1/x

take natural log on both sides

ln y = (1/x

Lim (1/x

x-->0

=Lim ln(cosx)/x

x-->0

=Lim -sinx/(2xcosx)

x-->0

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

x-->0

=-1/2

Therefore

lim (cosx)(1/x^2) = e

x-->0

- #4

- 191

- 0

Very nice...and simple...

Share: