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(B) To check the continuity at $x = 2$,we evaluate the left-hand limit,right-hand limit,and the value of the function at $x = 2$.$1$. Left-hand limit

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has an isolated point removable discontinuity

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(B) To check the continuity at $x = 2$,we evaluate the left-hand limit,right-hand limit,and the value of the function at $x = 2$.$1$. Left-hand limit $(LHL)$: $\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^-} \frac{x}{[x]}$. Since $1 \leqslant x $2$. Right-hand limit $(RHL)$: $\lim_{x 	o 2^+} f(x) = \lim_{x 	o 2^+} \sqrt{6-x} = \sqrt{6-2} = \sqrt{4} = 2$.$3$. Value of the function: $f(2) = 1$.Since $\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^+} f(x) = 2$,but $f(2) = 1$,the limit exists but is not equal to the function value. This type of discontinuity,where the limit exists but the point is defined elsewhere,is called a removable discontinuity (specifically,an isolated point removable discontinuity).

Consider the function $f(x) = \begin{cases} \frac{x}{[x]} & \text{if } 1 \leqslant x < 2 \\ 1 & \text{if } x = 2 \\ \sqrt{6-x} & \text{if } 2 < x \leqslant 3 \end{cases}$ where $[x]$ denotes the greatest integer function. At $x = 2$,the function:

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