If f(x) = begin{cases} (1+|sin x|)^{frac{a}{| | vedclass

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

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$2/3, e^{2/3}$

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(C) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$.First,calculate the left-hand limit:$\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-} (1+|\sin x|)^{\frac{a}{|\sin x|}}$.Let $u = |\sin x|$. As $x 	o 0$,$u 	o 0^+$. The limit becomes $\lim_{u 	o 0^+} (1+u)^{a/u} = e^a$.Next,calculate the right-hand limit:$\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} e^{\frac{	an 2x}{	an 3x}}$.Using the limit $\lim_{	heta 	o 0} \frac{	an k	heta}{	heta} = k$,we get $\lim_{x 	o 0^+} \frac{	an 2x}{	an 3x} = \lim_{x 	o 0^+} \frac{(	an 2x / 2x) \cdot 2x}{(	an 3x / 3x) \cdot 3x} = \frac{1 \cdot 2}{1 \cdot 3} = \frac{2}{3}$.Thus,$\lim_{x 	o 0^+} f(x) = e^{2/3}$.Since $f(0) = b$,for continuity,$e^a = e^{2/3} = b$.Therefore,$a = 2/3$ and $b = e^{2/3}$.

If $f(x) = \begin{cases} (1+|\sin x|)^{\frac{a}{|\sin x|}}, & -\pi/6 < x < 0 \\ b, & x = 0 \\ e^{\frac{\tan 2x}{\tan 3x}}, & 0 < x < \pi/6 \end{cases}$ is continuous at $x = 0$,find the values of $a$ and $b$.

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