Carilah nilai $\lim\limits_{x\to 2} \frac{x^2+2x-8}{x^2-x-2}!$

Carilah nilai

lim_{x \to 2} \frac{x^{2} + 2 x - 8}{x^{2} - x - 2}!

Carilah nilai $lim_{x \to 2} \frac{x^{2} + 2 x - 8}{x^{2} - x - 2}!$

$x^{2} + 2 x - 8 = (x + 4) (x - 2)$

$x^{2} - x - 2 = (x + 1) (x - 2)$

$lim_{x \to 2} \frac{x^{2} + 2 x - 8}{x^{2} - x - 2} = lim_{x \to 2} \frac{(x + 4) (x - 2)}{(x + 1) (x - 2)}$

$= lim_{x \to 2} \frac{x + 4}{x + 1}$

substitusi $x = 2 \to \frac{x + 4}{x + 1}$

$= \frac{2 + 4}{2 + 1} = \frac{6}{3} = 2$

Jadi, Nilai $lim_{x \to 2} \frac{x^{2} + 2 x - 8}{x^{2} - x - 2} = 2$

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