Q.Test the continuity of the following function at the point indicated against it: $f(x) = \dfrac{(27-2x)^{1/3} - 3}{9 - 3(243+5x)^{1/5}}$, for $x \ne 0$, $= 2$, for $x = 0$, at $x = 0$.

$f(x)=\dfrac{(27-2x)^{1/3}-3}{9-3(243+5x)^{1/5}}$ for $x\ne0$, $f(0)=2$.

Let $N(x)=(27-2x)^{1/3}-3$ and $D(x)=9-3(243+5x)^{1/5}$; both $N(0)=27^{1/3}-3=0$ and $D(0)=9-3(243)^{1/5}=9-3(3)=0$, a $0/0$ form. Treat each as a difference quotient by differentiating and evaluating at $x=0$ (equivalently, using the linear approximation near $x=0$):

$N'(x)=\tfrac13(27-2x)^{-2/3}(-2)=-\tfrac23(27-2x)^{-2/3},\qquad N'(0)=-\tfrac23\cdot27^{-2/3}=-\tfrac23\cdot\tfrac19=-\tfrac{2}{27}.$ …

Recognise both numerator and denominator as $0/0$ at $x=0$ and evaluate each as a derivative-style difference …

Losing track of the exponent laws when differentiating fractional powers (e.g. mis-simplifying $27^{-2/3}$ or $243^{-4/5}$) — it helps to note $27=3^3$ and …