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If $f(x)= \begin{cases}\frac{\sin (1+[x])}{[x]}, & \text { for }[x] \neq 0 \\ 0, & \text { for }[x]=0\end{cases}$
where $[x]$ denotes the greatest integer not exceeding $x$, then $\lim _{x \rightarrow 0^{-}} f(x)$ is equal to
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where $[x]$ denotes the greatest integer not exceeding $x$, then $\lim _{x \rightarrow 0^{-}} f(x)$ is equal to
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The correct answer is:
$0$
Given, $\begin{aligned} f(x) & = \begin{cases}\frac{\sin (1+[x])}{[x]}, & \text { for }[x] \neq 0 \\ 0 \quad & \text { for }[x]=0\end{cases} \\ \lim _{x \rightarrow 0^{-}} f(x) & =\lim _{x \rightarrow 0^{-}} \frac{\sin (1+[x])}{[x]} \\ & =\frac{\sin (1-1)}{-1} \\ & =0\end{aligned}$
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