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Question
If \[f\left( x \right) = \begin{cases}\frac{\sin\left[ x \right]}{\left[ x \right]}, & \left[ x \right] \neq 0 \\ 0, & \left[ x \right] = 0\end{cases}\] where denotes the greatest integer function, then \[\lim_{x \to 0} f\left( x \right)\]
Options
1
0
−1
does not exist
Solution
We have,
\[\left[ x \right] = \begin{cases}0, & 0 \leq x < 1 \\ - 1, & - 1 \leq x < 0\end{cases}\]
\[\therefore f\left( x \right) = \begin{cases}\frac{\sin\left( - 1 \right)}{- 1}, & - 1 \leq x < 0 \\ 0, & 0 \leq x < 1\end{cases}\]
\[ \Rightarrow f\left( x \right) = \begin{cases}\sin1, & - 1 \leq x < 0 \\ 0, & 0 \leq x < 1\end{cases}\]
Now,
\[\lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0} \sin1 = \sin1\]
\[ \lim_{x \to 0^+} f\left( x \right) = \lim_{x \to 0} 0 = 0\]
Clearly,
\[\lim_{x \to 0^-} f\left( x \right) \neq \lim_{x \to 0^+} f\left( x \right)\]
Thus,
\[\lim_{x \to 0} f\left( x \right)\]does not exist.
Hence, the correct answer is option (d).
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