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

• A

  1
• B

  0
• Correct answer
  C

  $-\cos k$
• D

  $\sin k$

Solution

We have, $\begin{aligned} & \lim _{x \rightarrow k} \frac{\sin \left(2 \pi\left(([x])-\left[\frac{x}{k}\right]\right)-x\right)+\sin k}{x-k} \\ & =\lim _{x \rightarrow k} \frac{\sin (2 \pi m-x)+\sin k}{x-k} \\ & {\left[\because[x]-\left[\frac{x}{k}\right]=m \text { is an integer }\right]} \\ \end{aligned}$ $=\lim _{x \rightarrow k} \frac{-\sin x+\sin k}{x-k}$
$=-\cos k$

Hence, the correct answer is option: 'C'.