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(C) For the function $f(x)$ to be continuous at $x=0$, we must have $\lim_{x 	o 0} f(x) = f(0)$.Given $f(x) = \frac{\sin [-x^2]}{[-x^2]}$ for $x 
eq

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$\alpha=\sin (1)$

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(C) For the function $f(x)$ to be continuous at $x=0$, we must have $\lim_{x 	o 0} f(x) = f(0)$.Given $f(x) = \frac{\sin [-x^2]}{[-x^2]}$ for $x 
eq 0$.As $x 	o 0$, $x^2$ approaches $0$ from the positive side, so $-x^2$ approaches $0$ from the negative side (i.e.,$-x^2 \in (-1, 0)$).Therefore, the greatest integer function $[-x^2]$ will take the value $-1$ as $x 	o 0$.Thus, $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \frac{\sin [-x^2]}{[-x^2]} = \frac{\sin(-1)}{-1} = \frac{-\sin(1)}{-1} = \sin(1)$.Since $f(0) = \alpha$, for continuity, we must have $\alpha = \sin(1)$.

Let $[x]$ denote the greatest integer less than or equal to $x$. Then, the value of $\alpha$ for which the function $f(x)=\begin{cases} \frac{\sin [-x^2]}{[-x^2]}, & x \neq 0 \\ \alpha, & x=0 \end{cases}$ is continuous at $x=0$ is:

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