Question

lim x tends to 2 (x^2-5x+6/x^2-4)​

Answer 1

Answer:

-0.25

Step-by-step explanation:

$lim_{x \: - > \: 2} \: \frac{x^2-5x+6}{x^2-4} \\ \\ lim_{x \: - > \: 2} \: \frac{x^2-5x+6}{(x - 2)(x + 2)} \\ \\ lim_{x \: - > \: 2} \: \frac{(x - 2)(x - 3)}{(x-2)(x + 2)} \\ \\ lim_{x \: - > \: 2} \: \frac{x - 3}{x + 2} \\ \\ \frac{2 - 3}{2 + 2} \\ \\ \frac{ - 1}{4} \\ \\ = > lim_{x \: - > \: 2} \: \frac{x^2-5x+6}{x^2-4} = - \frac{1}{4}$

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Answer 2

Answer:

Answer:-0.25

Step-by-step explanation:

$\begin{lgathered}lim_{x \: - > \: 2} \: \frac{x^2-5x+6}{x^2-4} \\ \\ lim_{x \: - > \: 2} \: \frac{x^2-5x+6}{(x - 2)(x + 2)} \\ \\ lim_{x \: - > \: 2} \: \frac{(x - 2)(x - 3)}{(x-2)(x + 2)} \\ \\ lim_{x \: - > \: 2} \: \frac{x - 3}{x + 2} \\ \\ \frac{2 - 3}{2 + 2} \\ \\ \frac{ - 1}{4} \\ \\ = > lim_{x \: - > \: 2} \: \frac{x^2-5x+6}{x^2-4} = - \frac{1}{4} \\ \\\end{lgathered}$