If f(x) = begin{cases} frac{sin [x]}{[x] + 1} | vedclass

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(A) For $f(x)$ to be continuous at $x = 0$,the condition $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$ must hold.First,calculate the left-han

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Equal to $0$

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(A) For $f(x)$ to be continuous at $x = 0$,the condition $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$ must hold.First,calculate the left-hand limit $(LHL)$ at $x = 0$:$\lim_{x 	o 0^-} f(x) = \lim_{h 	o 0} f(0 - h) = \lim_{h 	o 0} \frac{\cos \frac{\pi}{2}[-h]}{[-h]}$.Since $h > 0$ is a very small positive number,$[-h] = -1$.Thus,$\lim_{h 	o 0} \frac{\cos \frac{\pi}{2}(-1)}{-1} = \frac{\cos(-\frac{\pi}{2})}{-1} = \frac{0}{-1} = 0$.Next,calculate the right-hand limit $(RHL)$ at $x = 0$:$\lim_{x 	o 0^+} f(x) = \lim_{h 	o 0} f(0 + h) = \lim_{h 	o 0} \frac{\sin [h]}{[h] + 1}$.Since $h > 0$ is a very small positive number,$[h] = 0$.Thus,$\lim_{h 	o 0} \frac{\sin(0)}{0 + 1} = \frac{0}{1} = 0$.Since $	ext{LHL} = 	ext{RHL} = 0$,for the function to be continuous,$f(0) = k$ must equal $0$.

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