If for p ≠ q ≠ 0 , then function f(x) = (p(729 + x)7 - 3)/(729 +...

For a function to be continuous $f$\left(x\right)$to be$x = 0$, we know </p><p>$f\left(0\right) = \lim_{x \to 0}f\left(x\right)$</p><p>For$f\left(x\right) = \dfrac{p\left(729 + x\right)7 - 3}{729 + qx3 - 9}$</p><p>limit should be$\dfrac{0}{0}$from</p><p>So,$\lim_{x \to 0}\left(p\left(729 + x\right)7 - 3\right)$</p><p>$p . 7297 - 3 = 0$</p><p>$⇒ p \cdot 729 = {3}^{7} ⇒ p = 3$</p><p>Now,$f\left(0\right) = \lim_{x \to 0}\dfrac{3\left({3}^{6} + x\right)7 - 3}{{3}^{6} + qx3 - 9}$</p><p>$= \lim_{x \to 0}\dfrac{3\left[{\left(1 + \dfrac{x}{{3}^{6}}\right)}^{\dfrac{1}{7}} - 1\right]}{9\left[{\left(1 + \dfrac{qx}{{3}^{6}}\right)}^{\dfrac{1}{3}} - 1\right]} = \dfrac{3}{9} \times \dfrac{\dfrac{1}{7 \cdot {3}^{6}}}{\dfrac{q}{3 \cdot {3}^{6}}}$</p><p>$⇒ f\left(0\right) = \dfrac{1}{3} \times \dfrac{3}{7q} = \dfrac{1}{7q}$</p><p>$⇒ 7qf\left(0\right) - 1 = 0$</p><p>$⇒ 7 \cdot {p}^{2} \cdot qf\left(0\right) - {p}^{2} = 0$(multiplying the equation by${p}^{2}$)</p><p>$⇒ 63qf\left(0\right) - {p}^{2} = 0