JAMB - Mathematics (2023 - No. 16)
Evaluate the following limit: \(lim_{x\to2} \frac {x^2 + 4x - 12}{x^2 - 2x}\)
4
8
0
2
Giải thích
\(lim_{x\to2} \frac {x^2 + 4x - 12}{x^2 - 2x}\) = \(lim_{x\to2} \frac {(x - 2)(x + 6)}{x(x - 2)}\)
\(lim_{x\to2} \frac {x + 6}{x}\)
\(\frac {2 + 6}{2} = \frac {8}{2}\) = 4