Hi everybody,

I have two questions:

1)- $\displaystyle \lim_{x\to 0}\frac{sinKx}{x}= K ??? \$

2)-$\displaystyle \lim_{x\to 0} \frac{1-cosKx}{x^2}= \frac{K}{2} ???$

Thank you anyway.

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- Aug 12th 2009, 06:13 PM #1

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- Aug 12th 2009, 07:33 PM #2

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As before, use L'Hopital's rule. The derivative of sin(Kx) is K cos(Kx) and the derivative of x is 1. The limit is $\displaystyle \lim_{x\to 0}K cos(Kx)$.

The derivative of 1- cos(Kx) is K sin(Kx) and the derivative of $\displaystyle x^2$ is 2x. Since those both go to 0, do it again.