Let [t] denote the greatest integer leq t and | vedclass

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(B) Given $A = \lim_{x 	o 0} x[\frac{4}{x}]$.Using the property $[y] = y - \{y\}$,we have $A = \lim_{x 	o 0} x(\frac{4}{x} - \{\frac{4}{x}\}) = \lim

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$\sqrt{A+1}$

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(B) Given $A = \lim_{x 	o 0} x[\frac{4}{x}]$.Using the property $[y] = y - \{y\}$,we have $A = \lim_{x 	o 0} x(\frac{4}{x} - \{\frac{4}{x}\}) = \lim_{x 	o 0} (4 - x\{\frac{4}{x}\})$.Since $0 \leq \{\frac{4}{x}\} Thus,$A = 4$.The function $f(x) = [x^2] \sin(\pi x)$ is discontinuous where $[x^2]$ is discontinuous,which occurs when $x^2$ is an integer,excluding points where $\sin(\pi x) = 0$.For $x^2 = k$ (where $k \in \mathbb{Z}$),$f(x)$ is discontinuous unless $\sin(\pi \sqrt{k}) = 0$.Checking the options for $A=4$: $\sqrt{A+1} = \sqrt{5}$. Since $x^2 = 5$ is an integer and $\sin(\pi \sqrt{5}) 
eq 0$,the function is discontinuous at $x = \sqrt{5}$.

Let $[t]$ denote the greatest integer $\leq t$ and $\lim_{x \to 0} x[\frac{4}{x}] = A$. Then the function $f(x) = [x^2] \sin(\pi x)$ is discontinuous when $x$ is equal to

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