The function f(x) = (x - a)^2 cos frac{1}{(x- | vedclass

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(B) To check for continuity at $x = a$,we evaluate $\lim_{x 	o a} f(x)$:$\lim_{x 	o a} (x - a)^2 \cos \frac{1}{x-a}$.Since $-1 \leq \cos \frac{1}{x-

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derivable at $x = a$

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(B) To check for continuity at $x = a$,we evaluate $\lim_{x 	o a} f(x)$:$\lim_{x 	o a} (x - a)^2 \cos \frac{1}{x-a}$.Since $-1 \leq \cos \frac{1}{x-a} \leq 1$,we have $-(x-a)^2 \leq (x-a)^2 \cos \frac{1}{x-a} \leq (x-a)^2$.By the Squeeze Theorem,as $x 	o a$,both $-(x-a)^2$ and $(x-a)^2$ approach $0$,so $\lim_{x 	o a} f(x) = 0 = f(a)$. Thus,$f(x)$ is continuous at $x = a$.To check for differentiability at $x = a$,we calculate $f'(a) = \lim_{h 	o 0} \frac{f(a+h) - f(a)}{h}$:$f'(a) = \lim_{h 	o 0} \frac{(a+h-a)^2 \cos \frac{1}{a+h-a} - 0}{h} = \lim_{h 	o 0} \frac{h^2 \cos \frac{1}{h}}{h} = \lim_{h 	o 0} h \cos \frac{1}{h}$.Again,since $-1 \leq \cos \frac{1}{h} \leq 1$,we have $-|h| \leq h \cos \frac{1}{h} \leq |h|$.As $h 	o 0$,both $-|h|$ and $|h|$ approach $0$,so $f'(a) = 0$. Therefore,the function is derivable at $x = a$.

The function $f(x) = (x - a)^2 \cos \frac{1}{(x-a)}$ for $x \neq a$ and $f(a) = 0$,is

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