7
2
3
4
Correct answer is B
\lim \limits_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4}
\frac{(x - 2)(x^{2} + 3x - 2)}{x^{2} - 4} = \frac{(x - 2)(x^{2} + 3x - 2)}{(x - 2)(x + 2)}
= \frac{(x^{2} + 3x - 2)}{x + 2}
\therefore \lim \limits_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4} = \lim \limits_{x \to 2} \frac{x^{2} + 3x - 2}{x + 2}
= \frac{2^{2} + 3(2) - 2}{2 + 2}
= \frac{4 + 6 - 2}{4} = 2
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