If f(x) = begin{cases} frac{a}{2}(x - |x|), & | vedclass

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(A) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0^{-}} f(x) = f(0) = \lim_{x 	o 0^{+}} f(x)$.First,consider the left-hand limit:

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$a$ is any real value and $b$ is any real value

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(A) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0^{-}} f(x) = f(0) = \lim_{x 	o 0^{+}} f(x)$.First,consider the left-hand limit: $\lim_{x 	o 0^{-}} f(x) = \lim_{x 	o 0^{-}} \frac{a}{2}(x - |x|)$.Since $x This holds true for any real value of $a$.Next,consider the right-hand limit: $\lim_{x 	o 0^{+}} f(x) = \lim_{x 	o 0^{+}} bx^2 \sin \left( \frac{1}{x} ight)$.We know that $-1 \leq \sin \left( \frac{1}{x} ight) \leq 1$.Multiplying by $bx^2$,we get $-|bx^2| \leq bx^2 \sin \left( \frac{1}{x} ight) \leq |bx^2|$.By the Squeeze Theorem,as $x 	o 0$,$bx^2 \sin \left( \frac{1}{x} ight) 	o 0$.This holds true for any real value of $b$.Thus,$f(x)$ is continuous at $x = 0$ for any real values of $a$ and $b$.

If $f(x) = \begin{cases} \frac{a}{2}(x - |x|), & \text{for } x < 0 \\ 0, & \text{for } x = 0 \\ bx^2 \sin \left( \frac{1}{x} \right), & \text{for } x > 0 \end{cases}$ is continuous at $x = 0$,then

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