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
- A$-1$
- B$0$
- C$1$
- D$2$
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