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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- Jun 3rd 2006, 09:31 AMNichelle14Please check my work
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. - Jun 3rd 2006, 10:55 AMCaptainBlackQuote:

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 - Jun 3rd 2006, 12:11 PMtopsquarkQuote:

Originally Posted by**Nichelle14**

$\displaystyle f'(x) = \frac{cos [ sin ( sin [kx] ) ] cos ( sin[kx] ) cos[kx] k}{x}$ - $\displaystyle \frac{sin [ sin ( sin[kx] ) ]}{x^2}$

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:

$\displaystyle \lim_{x \to 0}sin(x) \to x - (1/6)x^3$

and keep terms to third order in x.

So the numerator of your limit becomes:

$\displaystyle \lim_{x \to 0}sin [ sin ( sin[kx] ) ] \to \lim_{x \to 0} sin [ sin ( kx - (1/6)k^3x^3 ) ] $ $\displaystyle \to \lim_{x \to 0}sin [ \Big(kx - (1/6)k^3x^3 \Big) - (1/6) \Big(kx - (1/6)k^3x^3 \Big)^3 ] $

$\displaystyle \to \lim_{x \to 0}sin [ kx - (1/6)k^3x^3 - (1/6)k^3x^3 ] $ (to third order in x)

$\displaystyle \to \lim_{x \to 0}sin [ kx - (2/6)k^3x^3 ] $

$\displaystyle \to \lim_{x \to 0}[ \Big( kx - (2/6)k^3x^3 \Big) - (1/6) \Big( kx - (2/6)k^3x^3 \Big)^3] $ $\displaystyle \to \lim_{x \to 0}[kx - (2/6)k^3x^3 - (1/6)k^3x^3] $ $\displaystyle \to \lim_{x \to 0}[kx - (1/2)k^3x^3] $

Thus

$\displaystyle \lim_{x \to 0} \frac{sin [ sin ( sin[kx] ) ]}{x} \to \lim_{x \to 0} \frac{kx - (1/2)k^3x^3}{x}$ $\displaystyle \to k $

-Dan - Jun 3rd 2006, 12:18 PMCaptainBlackQuote:

Originally Posted by**topsquark**

**separately**, and

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 - Jun 3rd 2006, 12:21 PMtopsquarkQuote:

Originally Posted by**CaptainBlack**

-Dan - Jun 3rd 2006, 01:39 PMNichelle14
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. - Jun 3rd 2006, 02:14 PMCaptainBlackQuote:

Originally Posted by**Nichelle14**

that means all of the cosines go to 1, and so the whole product goes to k.

RonL - Jun 3rd 2006, 02:45 PMNichelle14
Thanks. I see where I left off my brackets.

You are pretty smart.

I appreciate the help you have given me. :)