If f(x) = begin{cases} frac{sin([x])}{[x]}, & | vedclass

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(D) To find $\lim_{x 	o 0} f(x)$,we evaluate the left-hand limit $(LHL)$ and the right-hand limit $(RHL)$ at $x = 0$.For the $LHL$: $\lim_{x 	o 0^-}

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Does not exist

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(D) To find $\lim_{x 	o 0} f(x)$,we evaluate the left-hand limit $(LHL)$ and the right-hand limit $(RHL)$ at $x = 0$.For the $LHL$: $\lim_{x 	o 0^-} f(x)$. As $x 	o 0^-$,$x$ is slightly less than $0$,so $[x] = -1$.Thus,$f(x) = \frac{\sin(-1)}{-1} = \frac{-\sin(1)}{-1} = \sin(1)$.So,$\lim_{x 	o 0^-} f(x) = \sin(1)$.For the $RHL$: $\lim_{x 	o 0^+} f(x)$. As $x 	o 0^+$,$x$ is in the interval $[0, 1)$,so $[x] = 0$.According to the definition of $f(x)$,when $[x] = 0$,$f(x) = 0$.So,$\lim_{x 	o 0^+} f(x) = 0$.Since $\lim_{x 	o 0^-} f(x) = \sin(1)$ and $\lim_{x 	o 0^+} f(x) = 0$,the left-hand limit is not equal to the right-hand limit.Therefore,$\lim_{x 	o 0} f(x)$ does not exist.

If $f(x) = \begin{cases} \frac{\sin([x])}{[x]}, & \text{when } [x] \neq 0 \\ 0, & \text{when } [x] = 0 \end{cases}$ where $[x]$ is the greatest integer function,then $\lim_{x \to 0} f(x) = $

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