If f(x) is continuous at point x=0 where f(x) | vedclass

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(B) For $f(x)$ to be continuous at $x=0$,we must have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0) = \frac{2}{\log 2}$.Step $1$: Evaluate $\l

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$16, 32$

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(B) For $f(x)$ to be continuous at $x=0$,we must have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0) = \frac{2}{\log 2}$.Step $1$: Evaluate $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-} \frac{3 \sin x + 5 	an x}{a^x - 1}$.Dividing numerator and denominator by $x$,we get $\lim_{x 	o 0^-} \frac{3(\frac{\sin x}{x}) + 5(\frac{	an x}{x})}{\frac{a^x - 1}{x}} = \frac{3(1) + 5(1)}{\ln a} = \frac{8}{\ln a}$.Equating to $f(0)$: $\frac{8}{\ln a} = \frac{2}{\ln 2} \implies \ln a = 4 \ln 2 = \ln(2^4) = \ln 16 \implies a = 16$.Step $2$: Evaluate $\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} \frac{8x + 2x \cos x}{b^x - 1}$.Dividing numerator and denominator by $x$,we get $\lim_{x 	o 0^+} \frac{8 + 2 \cos x}{\frac{b^x - 1}{x}} = \frac{8 + 2(1)}{\ln b} = \frac{10}{\ln b}$.Equating to $f(0)$: $\frac{10}{\ln b} = \frac{2}{\ln 2} \implies \ln b = 5 \ln 2 = \ln(2^5) = \ln 32 \implies b = 32$.Thus,the values are $a=16$ and $b=32$.

If $f(x)$ is continuous at point $x=0$ where $f(x) = \begin{cases} \frac{3 \sin x + 5 \tan x}{a^x - 1} & , x < 0 \\ \frac{2}{\log 2} & , x = 0 \\ \frac{8x + 2x \cos x}{b^x - 1} & , x > 0 \end{cases}$ then the values of $a$ and $b$,respectively,are

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