JAMB 1992 · UME · Q36

Evaluate $\displaystyle\lim_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4}$ .

Worked solution (try it first)

Putting in $x = 2$ straight away gives $\frac00$ , so simplify first.
Factorise the bottom as a difference of two squares: $x^2 - 4 = (x - 2)(x + 2)$ .
Cancel the common factor $x - 2$ : the expression becomes $\dfrac{x^2 + 3x - 2}{x + 2}$ .
Now put in $x = 2$ : $\dfrac{4 + 6 - 2}{4} = \dfrac84 = 2$ , option B.