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Let $[x]$ denote the greatest integer less than or equal to $x$ and $k \geq 2$ be an integer. Then $\operatorname{Lt}_{x \rightarrow k} \frac{\sin \left(2 \pi\left([x]-\left[\frac{x}{k}\right]\right)-x\right)+\sin k}{x-k}=$
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Verified Answer
The correct answer is:
$-\cos k$
We have,
$\begin{aligned}
& \lim _{x \rightarrow k} \frac{\sin \left(2 \pi\left(([x])-\left[\frac{x}{k}\right]\right)-x\right)+\sin k}{x-k} \\
& =\lim _{x \rightarrow k} \frac{\sin (2 \pi m-x)+\sin k}{x-k} \\
& {\left[\because[x]-\left[\frac{x}{k}\right]=m \text { is an integer }\right]} \\
\end{aligned}$
$=\lim _{x \rightarrow k} \frac{-\sin x+\sin k}{x-k}$
$=-\cos k$
$\begin{aligned}
& \lim _{x \rightarrow k} \frac{\sin \left(2 \pi\left(([x])-\left[\frac{x}{k}\right]\right)-x\right)+\sin k}{x-k} \\
& =\lim _{x \rightarrow k} \frac{\sin (2 \pi m-x)+\sin k}{x-k} \\
& {\left[\because[x]-\left[\frac{x}{k}\right]=m \text { is an integer }\right]} \\
\end{aligned}$
$=\lim _{x \rightarrow k} \frac{-\sin x+\sin k}{x-k}$
$=-\cos k$
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