If for p neq q neq 0,the function f(x) = frac | vedclass

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(B) Since $f(x)$ is continuous at $x=0$,we have $f(0) = \lim_{x 	o 0} f(x)$.For the limit to exist,it must be in the $\frac{0}{0}$ form.Substituting

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$63qf(0)-p^2=0$

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(B) Since $f(x)$ is continuous at $x=0$,we have $f(0) = \lim_{x 	o 0} f(x)$.For the limit to exist,it must be in the $\frac{0}{0}$ form.Substituting $x=0$ in the numerator: $\sqrt[7]{p(729)} - 3 = 0 \implies p(3^6) = 3^7 \implies p = 3$.Now,$f(0) = \lim_{x 	o 0} \frac{\sqrt[7]{3(3^6+x)}-3}{\sqrt[3]{3^6+qx}-9} = \lim_{x 	o 0} \frac{3[(1+\frac{x}{3^6})^{1/7}-1]}{9[(1+\frac{qx}{3^6})^{1/3}-1]}$.Using the standard limit $\lim_{u 	o 0} \frac{(1+u)^n-1}{u} = n$,we get:$f(0) = \frac{3}{9} 	imes \frac{\frac{1}{7} \cdot \frac{1}{3^6}}{\frac{1}{3} \cdot \frac{q}{3^6}} = \frac{1}{3} 	imes \frac{3}{7q} = \frac{1}{7q}$.Thus,$7qf(0) = 1$,which implies $7qf(0) - 1 = 0$.Since $p=3$,$p^2 = 9$. Substituting $1 = \frac{p^2}{9}$ into the equation:$7qf(0) - \frac{p^2}{9} = 0 \implies 63qf(0) - p^2 = 0$.

If for $p \neq q \neq 0$,the function $f(x) = \frac{\sqrt[7]{p(729+x)}-3}{\sqrt[3]{729+qx}-9}$ is continuous at $x=0$,then:

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