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

<p>$f(x) = {{\root 7 \of {p(729 + x)} - 3} \over {\root 3 \of {729 + qx} - 9}}$</p> <p>for continuity at $x = 0$, $\mathop {\lim }\limits_{x \to 0} f(x) = f(0)$</p> <p>Now, $\therefore$ $$\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} {{\root 7 \of {p(729 + x)} - 3} \over {\root 3 \of {729 + qx} - 9}}$$</p> <p>$\Rightarrow p = 3$ (To make indeterminant form)</p> <p>So, $$\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} {{{{\left( {{3^7} + 3x} \right)}^{{1 \over 7}}} - 3} \over {{{\left( {729 + qx} \right)}^{{1 \over 3}}} - 9}}$$</p> <p>$$ = \mathop {\lim }\limits_{x \to 0} {{3\left[ {{{\left( {1 + {x \over {{3^6}}}} \right)}^{{1 \over 7}}} - 1} \right]} \over {9\left[ {{{\left( {1 + {q \over {729}}x} \right)}^{{1 \over 3}}} - 1} \right]}} = {1 \over 3}\,.\,{{{1 \over 7}\,.\,{1 \over {{3^6}}}} \over {{1 \over 3}\,.\,{q \over {729}}}}$$</p> <p>$\therefore$ $f(0) = {1 \over {7q}}$</p> <p>$\therefore$ Option (B) is correct.</p>