Question

$\displaystyle \lim_{x \to 2} \frac{x^2-4}{x-2}$
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Answer 1

Answer:

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

Answer 2

$= \displaystyle \lim_{x \to 2} \frac{x^2-4}{x-2}$

$= \displaystyle \lim_{x \to 2} \frac{x^2- {2}^{2} }{x-2}$

$= \displaystyle \lim_{x \to 2} \frac{(x + 2)(x - 2) }{x-2}$

$= \displaystyle \lim_{x \to 2} \frac{(x + 2) \cancel{(x - 2)} }{ \cancel{x-2}}$

$= \displaystyle \lim_{x \to 2} (x + 2)$

$= (2 + 2)$

$= 4$