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(C) Let $f(x) = \sin \left(2 \pi\left([x]-\left[\frac{x}{k}\right]\right)-x\right) + \sin k$. As $x \rightarrow k$,the expression $[x] - [x/k]$ takes

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$-\cos k$

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(C) Let $f(x) = \sin \left(2 \pi\left([x]-\left[\frac{x}{k}\right]\right)-x\right) + \sin k$. As $x \rightarrow k$,the expression $[x] - [x/k]$ takes integer values. Specifically,for $x$ in a small neighborhood of $k$,$[x] = k$ and $[x/k] = 1$ (since $k \geq 2$). Thus,the term $2 \pi ([x] - [x/k])$ is an even multiple of $\pi$,say $2 \pi m$. So,$\sin(2 \pi m - x) = \sin(-x) = -\sin x$. The limit becomes $\lim_{x \rightarrow k} \frac{-\sin x + \sin k}{x - k}$. Applying $L$'$H$ôpital's rule,we differentiate the numerator and denominator with respect to $x$: $\lim_{x \rightarrow k} \frac{-\cos x}{1} = -\cos k$.

Let $[x]$ denote the greatest integer less than or equal to $x$ and $k \geq 2$ be an integer. Then $\lim_{x \rightarrow k} \frac{\sin \left(2 \pi\left([x]-\left[\frac{x}{k}\right]\right)-x\right)+\sin k}{x-k} = $

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