If f(x) = begin{cases} frac{(e^{ax}-1) log(1+ | vedclass

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(A) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^-} f(x) = f(0) = 2$. First,evaluate the right-hand limi

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$4$

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(A) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^-} f(x) = f(0) = 2$. First,evaluate the right-hand limit $(RHL)$: $\lim_{x 	o 0^+} \frac{(e^{ax}-1) \log(1+x)}{\sin^2 x} = \lim_{x 	o 0^+} \frac{(\frac{e^{ax}-1}{ax} \cdot ax) \cdot (\frac{\log(1+x)}{x} \cdot x)}{(\frac{\sin x}{x})^2 \cdot x^2} = \lim_{x 	o 0^+} \frac{ax \cdot x}{x^2} = a$. Since $f(0) = 2$,we have $a = 2$. Next,evaluate the left-hand limit $(LHL)$: $\lim_{x 	o 0^-} \frac{\cos 4x - \cos bx}{	an^2 x} = \lim_{x 	o 0^-} \frac{-2 \sin(\frac{4x+bx}{2}) \sin(\frac{4x-bx}{2})}{	an^2 x} = \lim_{x 	o 0^-} \frac{-2 (\frac{4+b}{2}x) (\frac{4-b}{2}x)}{x^2} = -2 \cdot \frac{16-b^2}{4} = \frac{b^2-16}{2}$. Since $f(0) = 2$,we have $\frac{b^2-16}{2} = 2$,which implies $b^2 - 16 = 4$,so $b^2 = 20$. Finally,calculate $\sqrt{b^2 - a^2} = \sqrt{20 - 2^2} = \sqrt{20 - 4} = \sqrt{16} = 4$.

If $f(x) = \begin{cases} \frac{(e^{ax}-1) \log(1+x)}{\sin^2 x} & \text{if } x > 0 \\ 2 & \text{if } x = 0 \\ \frac{\cos 4x - \cos bx}{\tan^2 x} & \text{if } x < 0 \end{cases}$ is continuous at $x = 0$,then $\sqrt{b^2 - a^2} = $

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