find the lim as x approaches 0 for sin(sin(sin(kx)))/x
I used l'Hopital's rule and got this
lim as x approaches 0 for cos(sin(sin(kx)cos(sin(kx))kcos(kx)
the answer is 0.
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find the lim as x approaches 0 for sin(sin(sin(kx)))/x
I used l'Hopital's rule and got this
lim as x approaches 0 for cos(sin(sin(kx)cos(sin(kx))kcos(kx)
the answer is 0.
If you add brackets so that they are properly balanced your limit looksQuote:
Originally Posted by Nichelle14
OK, but to me this looks like it goes to k.
Also numerical checks indicates the k is the right answer.
RonL
The reason this didn't work is that you are missing a term in the above. Let thenQuote:
Originally Posted by Nichelle14

This expression gives a mess if you try to take the limit as x goes to 0. (You wind up doing an infinite number of L'Hopital rules.)
My suggestion is to use a nested series of approximations for small values of x in the sine function:
and keep terms to third order in x.
So the numerator of your limit becomes:
(to third order in x)
Thus
Dan
In L'Hopitals rule you differentiate the top and bottom separately, andQuote:
Originally Posted by topsquark
the limit of the indeterminate form is the limit of the ratio of the derivatives.
As far as I can see the OP has found the ratio of derivatives correctly.
If you put the missing brackets in you will see that what she has does
in fact go to k as x goes to zero, as expected.
RonL
Oh dear, I'm blushing now. (Ahem!) I knew that. Really I did!Quote:
Originally Posted by CaptainBlack
Dan
I orginally thought it would be k, but couldn't prove my work.
I still don't see where I am going wrong. Can you explain a little more? I appreciate your help.
You have a product of three cosines of things which go to zero and k, butQuote:
Originally Posted by Nichelle14
that means all of the cosines go to 1, and so the whole product goes to k.
RonL
Thanks. I see where I left off my brackets.
You are pretty smart.
I appreciate the help you have given me. :)