WAEC 2022 · Paper 1 · Q38

Evaluate: $\displaystyle\lim_{x \to -2}\left(\frac{x^3 + 8}{x + 2}\right)$ .

Worked solution (try it first)

Putting $x = -2$ straight in gives $\frac00$ , so factorise the sum of cubes first: $x^3 + 8 = (x + 2)(x^2 - 2x + 4)$ .
Cancel $(x + 2)$ : the expression is $x^2 - 2x + 4$ for $x \ne -2$ .
Now put $x = -2$ : $4 + 4 + 4 = 12$ , option D.