The value of p for which the function f(x) = | vedclass

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(D) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0} f(x) = f(0) = 12(\log 4)^3$.First,we evaluate the limit:$\lim_{x 	o 0} f(x) =

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$4$

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(D) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0} f(x) = f(0) = 12(\log 4)^3$.First,we evaluate the limit:$\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \frac{(4^x - 1)^3}{\sin(\frac{x}{p}) \log(1 + \frac{x^2}{3})}$Rewrite the expression to use standard limits $\lim_{u 	o 0} \frac{a^u - 1}{u} = \log a$,$\lim_{u 	o 0} \frac{\sin u}{u} = 1$,and $\lim_{u 	o 0} \frac{\log(1+u)}{u} = 1$:$\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \left[ \left( \frac{4^x - 1}{x} ight)^3 \cdot x^3 \cdot \frac{1}{\sin(\frac{x}{p})} \cdot \frac{1}{\log(1 + \frac{x^2}{3})} ight]$Multiply and divide by $(\frac{x}{p})$ and $(\frac{x^2}{3})$:$= \lim_{x 	o 0} \left[ \left( \frac{4^x - 1}{x} ight)^3 \cdot \frac{x/p}{\sin(x/p)} \cdot p \cdot \frac{x^2/3}{\log(1 + x^2/3)} \cdot \frac{3}{x^2} \cdot x^3 ight]$$= (\log 4)^3 \cdot 1 \cdot p \cdot 1 \cdot 3 = 3p(\log 4)^3$Equating this to $f(0)$:$3p(\log 4)^3 = 12(\log 4)^3$$3p = 12 \implies p = 4$.

The value of $p$ for which the function $f(x) = \begin{cases} \frac{(4^x - 1)^3}{\sin(\frac{x}{p}) \log(1 + \frac{x^2}{3})}, & x \ne 0 \\ 12(\log 4)^3, & x = 0 \end{cases}$ is continuous at $x = 0$ is:

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