If f is defined by f(x) = begin{cases} frac{1 | vedclass

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(A) For $f$ to be continuous at $x=0$,we must have $\lim_{x 	o 0} f(x) = f(0)$.Given $f(0) = \frac{1}{2}$.Now,calculate the limit: $\lim_{x 	o 0} \f

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(A) For $f$ to be continuous at $x=0$,we must have $\lim_{x 	o 0} f(x) = f(0)$.Given $f(0) = \frac{1}{2}$.Now,calculate the limit: $\lim_{x 	o 0} \frac{1-\cos ax}{x \sin x}$.Using the identity $1-\cos ax = 2 \sin^{2}(\frac{ax}{2})$,the limit becomes $\lim_{x 	o 0} \frac{2 \sin^{2}(\frac{ax}{2})}{x \sin x}$.Divide numerator and denominator by $x^{2}$:$\lim_{x 	o 0} \frac{2 \frac{\sin^{2}(ax/2)}{x^{2}}}{\frac{\sin x}{x}} = \lim_{x 	o 0} \frac{2 \cdot (\frac{a}{2})^{2} \cdot (\frac{\sin(ax/2)}{ax/2})^{2}}{\frac{\sin x}{x}}$.Since $\lim_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$,the limit is $\frac{2 \cdot (a^{2}/4)}{1} = \frac{a^{2}}{2}$.Equating this to $f(0)$,we get $\frac{a^{2}}{2} = \frac{1}{2}$,which implies $a^{2} = 1$.

If $f$ is defined by $f(x) = \begin{cases} \frac{1-\cos ax}{x \sin x}, & x \neq 0 \\ \frac{1}{2}, & x = 0 \end{cases}$ and $f$ is continuous at $x=0$,then $a^{2} =$ . . . . . . .

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