If f(x) =<span class="math-tex">\(\left\lbrace \begin{matrix}\dfrac{\sin [\rm x]}{[\rm x]}, \ \ [\rm x] \neq 0 \\\ 0, \ \ [\rm x] = 0\end{matrix} \right.\)</span>, where [x] is the Greatest Integer but not larger than x, then <span class="math-tex">\(\rm \displaystyle\lim_{x \rightarrow 0} f(x)\)</span> is
As, \[ f(x) = \begin{cases} \frac{\sin[x]}{[x]}, & [x] \neq 0 \\ 0, & [x] = 0 \end{cases} \] \[ \Rightarrow f(x) = \begin{cases} \frac{\sin[x]}{[x]}, & x \in \mathbb{R} - [0,1) \\ 0, & 0 \leq x < 1 \end{cases} \] Right-Hand Limit (RHL) at \( x = 0 \): \[ \lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} \frac{\sin [0 + h]}{[0 + h]} = 0 \] Left-Hand Limit (LHL) at \( x = 0 \): \[ \lim\limits_{x \to 0^-} f(x) = \lim\limits_{h \to 0} \frac{\sin [0 - h]}{[0 - h]} \] \[ = \lim\limits_{h \to 0} \frac{\sin(-1)}{-1} = \sin 1 \] Since \( \text{RHL} \neq \text{LHL} \), Therefore, the limit does not exist.
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