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

Vedclass · Accepted Answer

$a = 2 / 3, b = e^{2 / 3}$

Vedclass · Answer

(B) For the function $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x ightarrow 0^-} f(x) = f(0) = \lim_{x ightarrow 0^+} f(x)$.First,calculate the left-hand limit:$\lim_{x ightarrow 0^-} f(x) = \lim_{x ightarrow 0^-} (1 + |\sin x|)^{a/|\sin x|}$.Since this is of the form $1^\infty$,we use the formula $\lim_{x ightarrow c} (1 + g(x))^{h(x)} = e^{\lim_{x ightarrow c} g(x)h(x)}$.$\lim_{x ightarrow 0^-} f(x) = e^{\lim_{x ightarrow 0} |\sin x| \cdot \frac{a}{|\sin x|}} = e^a$.Next,calculate the right-hand limit:$\lim_{x ightarrow 0^+} f(x) = \lim_{x ightarrow 0^+} e^{\frac{	an 2x}{	an 3x}} = e^{\lim_{x ightarrow 0} \frac{	an 2x}{2x} \cdot \frac{3x}{	an 3x} \cdot \frac{2x}{3x}} = e^{1 \cdot 1 \cdot 2/3} = e^{2/3}$.Since $f(0) = b$,we equate the limits:$e^a = b = e^{2/3}$.Therefore,$a = 2/3$ and $b = e^{2/3}$.

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

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