- #1

KLscilevothma

- 322

- 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

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- Thread starter KLscilevothma
- Start date

- #1

KLscilevothma

- 322

- 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

bogdan

- 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

KLscilevothma

- 322

- 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

bogdan

- 191

- 0

Very nice...and simple...

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