Q.Evaluate $\lim_{x \to 3} \dfrac{x^3-27}{x-3}$.

Substituting $x=3$ directly gives $\dfrac{0}{0}$, an indeterminate form, so the numerator must first be factorised to remove the common factor causing it.

Using $x^3-27 = (x-3)(x^2+3x+9)$, the $(x-3)$ factor cancels with the denominator, leaving $x^2+3x+9$, which equals $27$ at $x=3$.

Direct substitution of $x=3$ gives $\dfrac{3^3-27}{3-3} = \dfrac{0}{0}$, an indeterminate form, so factorisation is needed before a value can be found.

Using the identity $x^3-a^3 = (x-a)(x^2+ax+a^2)$ with $a=3$: $x^3 - 27 = (x-3)(x^2+3x+9)$

Substituting into the original expression: $\dfrac{x^3-27}{x-3} = \dfrac{(x-3)(x^2+3x+9)}{x-3} = x^2+3x+9, \quad x \neq 3$

(the cancellation is valid because a limit only considers $x$ near $3$, never $x=3$ itself). Now direct substitution applies: $\lim_{x \to 3} (x^2+3x+9) = 3^2+3(3)+9 = 9+9+9 = 27$

Cross-check using the standard power limit (covered later in this chapter), $\lim_{x \to a}\frac{x^n-a^n}{x-a} = na^{n-1}$: with $n=3, a=3$, this gives $3 \times 3^2 = 27$ — the same answer by an independent formula.

Numerical cross-check: at $x=3.001$, $\dfrac{x^3-27}{x-3} = \dfrac{27.027009-27}{0.001} \approx \dfrac{0.027009}{0.001} = 27.009$, closing in on $27$ as $x \to 3$, confirming the algebraic result.

Alternatively, quote the standard limit $\lim_{x \to a}\frac{x^n-a^n}{x-a}=na^{n-1}$ directly with $n=3,\ a=3$: $3(3)^2=27$.

A frequent error is cancelling $x^3$ and $27$ against $x$ and $3$ term-by-term without factorising properly — factorisation must use the correct identity $x^3-a^3=(x-a)(x^2+ax+a^2)$, not a term-by-term cancellation, which is not a valid algebraic operation.