WAEC Further Mathematics 2014 Paper

Concept:Factorise the numerator and denominator, then cancel the common factor causing the indeterminate form.Explanation:Substituting $x = 3$ directly gives $\frac{0}{0}$, so the expression must be simplified first.The denominator is $x^{2} - 9 = (x - 3)(x + 3)$.The numerator $x^{3} + x^{2} - 12x$ is factorised by taking $x$ common:$x(x^{2} + x - 12) = x(x + 4)(x - 3)$.Therefore, the limit becomes:$\lim\limits_{x \to 3} \frac{x(x + 4)(x - 3)}{(x - 3)(x + 3)}$.Cancel $(x - 3)$, valid since $x \neq 3$ while taking the limit:$\lim\limits_{x \to 3} \frac{x(x + 4)}{x + 3}$.Now substitute $x = 3$:$\frac{3(3 + 4)}{3 + 3} = \frac{3 \times 7}{6} = \frac{21}{6} = \frac{7}{2}$.Answer:$\frac{7}{2}$Thus, the correct option is A.