If lim_{x to 0} frac{(4^x - 1)^3}{tan(frac{x} | vedclass

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(A) We use the standard limits: $\lim_{x 	o 0} \frac{a^x - 1}{x} = \log a$, $\lim_{x 	o 0} \frac{	an x}{x} = 1$, and $\lim_{x 	o 0} \frac{\log(1+x

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$5$

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(A) We use the standard limits: $\lim_{x 	o 0} \frac{a^x - 1}{x} = \log a$, $\lim_{x 	o 0} \frac{	an x}{x} = 1$, and $\lim_{x 	o 0} \frac{\log(1+x)}{x} = 1$.Divide numerator and denominator by $x^3$:$\lim_{x 	o 0} \frac{(\frac{4^x - 1}{x})^3}{\frac{	an(x/4)}{x} \cdot \frac{\log(1 + x^2/3)}{x^2} \cdot x} = \lim_{x 	o 0} \frac{(\log 4)^3}{\frac{1}{4} \cdot \frac{1}{3} \cdot x} = \infty$.Wait, re-evaluating the expression: $\frac{(4^x - 1)^3}{	an(x/4) \log(1 + x^2/3)} \approx \frac{(x \log 4)^3}{(x/4) \cdot (x^2/3)} = \frac{x^3 (\log 4)^3}{x^3 / 12} = 12 (\log 4)^3$.Given $12 (\log 4)^3 = 96 (\log a)^b$, we have $12 (\log 4)^3 = 12 \cdot 8 (\log a)^b = 12 (2 \log 4)^3 = 12 (\log 4^2)^3 = 12 (\log 16)^3$.Comparing $12 (\log 4)^3 = 96 (\log a)^b$, we rewrite $12 (\log 4)^3 = 96 (\frac{\log 4}{2})^3 = 96 (\log 4^{1/2})^3 = 96 (\log 2)^3$.Thus, $a = 2, b = 3$. Therefore, $a + b = 2 + 3 = 5$.

If $\lim_{x \to 0} \frac{(4^x - 1)^3}{\tan(\frac{x}{4}) \log(1 + \frac{x^2}{3})} = 96(\log a)^b$, then $(a + b) = $

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