Let f: left(-frac{pi}{4}, frac{pi}{4}right) r | vedclass

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(B) For $f(x)$ to be continuous at $x=0$,we must have $\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x) = f(0) = b$.First,calculate the r

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$1+e$

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(B) For $f(x)$ to be continuous at $x=0$,we must have $\lim_{x ightarrow 0^-} f(x) = \lim_{x ightarrow 0^+} f(x) = f(0) = b$.First,calculate the right-hand limit:$\lim_{x ightarrow 0^+} f(x) = \lim_{x ightarrow 0^+} e^{\frac{\cot 4x}{\cot 2x}} = \lim_{x ightarrow 0^+} e^{\frac{	an 2x}{	an 4x}} = \lim_{x ightarrow 0^+} e^{\frac{	an 2x}{2	an 2x(1-	an^2 2x)}} = e^{1/2}$.Thus,$b = e^{1/2}$.Next,calculate the left-hand limit:$\lim_{x ightarrow 0^-} f(x) = \lim_{x ightarrow 0^-} (1+|\sin x|)^{\frac{3a}{|\sin x|}} = \lim_{t ightarrow 0^+} (1+t)^{\frac{3a}{t}} = e^{3a}$ (where $t = |\sin x|$).Equating the limits,$e^{3a} = e^{1/2}$,which implies $3a = 1/2$,so $a = 1/6$.Therefore,$6a = 1$.Finally,$6a + b^2 = 1 + (e^{1/2})^2 = 1 + e$.

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