Q.Evaluate $\lim_{x \to 1} \dfrac{x^3-1}{x^2-1}$.

Direct substitution gives $\dfrac00$, so the indeterminate form must be resolved by factorising both numerator and denominator before cancelling.

Factorising gives $\dfrac{(x-1)(x^2+x+1)}{(x-1)(x+1)}$; cancelling $(x-1)$ leaves $\dfrac{x^2+x+1}{x+1}$, which equals $\dfrac32$ at $x=1$.

Substituting $x=1$ directly gives $\dfrac{1-1}{1-1} = \dfrac00$, an indeterminate form, so both numerator and denominator must be factorised before the limit can be evaluated.

Using $x^3-1=(x-1)(x^2+x+1)$ and $x^2-1=(x-1)(x+1)$: $\dfrac{x^3-1}{x^2-1} = \dfrac{(x-1)(x^2+x+1)}{(x-1)(x+1)} = \dfrac{x^2+x+1}{x+1}, \quad x \neq 1$

Now direct substitution applies: $\lim_{x \to 1} \dfrac{x^2+x+1}{x+1} = \dfrac{1+1+1}{1+1} = \dfrac{3}{2}$

Numerical cross-check: at $x=1.001$, numerator $=1.001^3-1 \approx 0.003003$, denominator $=1.001^2-1 \approx 0.002001$, giving a ratio of $\approx 1.5007$, closing in on $\dfrac32=1.5$, confirming the algebraic result.

Students sometimes cancel the exponents (treating $x^3/x^2$ as if the $-1$s simply disappear) instead of correctly factorising the difference-of-cubes and difference-of-squares identities separately — each must be factorised on its own before any cancellation is valid.