If $\begin{cases}\frac{\sin[\text{x}]}{[\text{x}]} & \text{x}\neq0\\0, & [\text{x}]=0\end{cases}$ where [.] denotes the greatest integer function. then $\lim\limits_{\text{x} \rightarrow 0}\text{f}(\text{x})$ is equal to :
Solution:
Given $\begin{cases}\frac{\sin[\text{x}]}{[\text{x}]} & \text{x}\neq0\\0, & [\text{x}]=0\end{cases}$
$\text{L}.\text{H}.\text{H}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin[\text{x}]}{[\text{x}]}=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin[0-\text{h}]}{[0-\text{h}]} $
$=\lim\limits_{\text{h} \rightarrow 0}\frac{-\sin[\text{-h}]}{[-\text{h}]}=-1$
$\text{R}.\text{H}.\text{L}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin[\text{x}]}{[\text{x}]}=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin[0+\text{h}]}{[0+\text{h}]}=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin[\text{h}]}{[\text{h}]}=1$
$\text{L}.\text{H}.\text{L}\neq\text{R}.\text{H}.\text{L}$
So, the limit does not exist.
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