Q.Evaluate $\displaystyle\lim_{x\to3}\frac{x^5-243}{x-3}$ using the standard algebraic limit formula.

Direct substitution gives $\tfrac00$, so this matches the pattern $\dfrac{x^n-a^n}{x-a}$ with $n=5,\,a=3$ (since $3^5=243$).\n> [!TLDR]\n> Match $n=5$, $a=3$; apply …

Direct substitution of $x=3$ gives $\dfrac{3^5-243}{3-3}=\dfrac{243-243}{0}=\dfrac00$ — an indeterminate form, so we cannot substitute directly; the standard formula of §4 is designed exactly for this.

Recognise the pattern: $\dfrac{x^5-243}{x-3}$ has the form $\dfrac{x^n-a^n}{x-a}$ with $n=5$ and $a=3$ (checking: $a^n = 3^5=243$ ✓, matching the numerator's constant term).

Apply the standard formula $\lim_{x\to a}\dfrac{x^n-a^n}{x-a}=na^{n-1}$:

$\lim_{x\to3}\frac{x^5-243}{x-3} = 5\cdot 3^{4} = 5\cdot81 = 405.$ …

Algebraic factoring cross-check (shown in full above): factoring $x^5-3^5$ as $(x-3)(x^4+3x^3+9x^2+27x+81)$ and substituting $x=3$ into the simplified quoti …

A frequent slip is computing $na^{n-1}$ as $5\cdot3^{5}$ (forgetting to reduce the exponent by one) instead of $5\cdot3^{4}$ — always do …