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(D) We are given the displacement function $s(t) = \frac{A}{c^2 - k^2} (\sin kt - \sin ct)$.To find the limiting value as $c 	o k$,we evaluate $\lim_

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$\frac{-At \cos kt}{2k}$

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(D) We are given the displacement function $s(t) = \frac{A}{c^2 - k^2} (\sin kt - \sin ct)$.To find the limiting value as $c 	o k$,we evaluate $\lim_{c 	o k} \frac{A(\sin kt - \sin ct)}{c^2 - k^2}$.Since this is a $0/0$ form,we apply $L$'$H$ôpital's Rule with respect to $c$:$\lim_{c 	o k} \frac{\frac{d}{dc} [A(\sin kt - \sin ct)]}{\frac{d}{dc} [c^2 - k^2]} = \lim_{c 	o k} \frac{A(0 - t \cos ct)}{2c}$.Substituting $c = k$ into the expression,we get:$\frac{A(-t \cos kt)}{2k} = \frac{-At \cos kt}{2k}$.

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