Was given the function: limit as x approaches 0 of (cosx)^cscx
took the ln of both sides etc etc and worked it out to the lny=-sin/cos^2x = -0/1

the the last step they do shows lim as x approaches 0 of my beginning function= lim as x approaches 0 of e^lny=1

why do they do that? and why does it = 1?

im sure the answer is simple but my mind is burnt out

ok first you would just find the series expansion about 0 and show its 1..but for your case

lny=0
e^lny=e^0=y=1?

?

well after taking the ln, it works on to the lny=-0/1=0 but the original function wasnt an ln

so would i pretty much plug in values close to 0 and plug in for lny since the ln of 0 doesnt exist? and then take the e of that to get rid of the original ln i used?

using the following identity from http://en.wikipedia.org/wiki/Indeterminate%5Fform



I did this:


This post was edited by Azrad on Oct 25 2013 04:40pm

what

ln y is 0 what is y?
it's not ln 0
ln(1)=0

get some sleep

ha yeah i think i need to. This should be the easiest thing ive done all day but i suppose its just one of those things that im overthinking because ive been at it so long

With Landau notations, when x approaches zero :

cos (x) = 1 - x²/2 + O(x^4)

ln (cos(x)) = -x²/2 + O(x^4)

sin (x) = x + O(x^3) = x ( 1 + O(x²) )

1/sin(x) = 1/x * 1 / ( 1+O(x²) ) = 1/x * ( 1 + O(x²) ) = 1/x + O(x)

cos(x) ^ (1/sin(x)) = exp ( 1/sin(x) * ln(cos(x)) )

= exp ( (1/x + O(x)) * ( -x²/2 + O(x^4) )

= exp ( -x/2 + O(x^3) )

Conclusion :

cos(x)^csc(x) ~ exp (-x/2)

as x approaches zero.

thanks to both of you. After some well needed mental rest i get it. Now to move on to improper integrals