Question

Va rog cu limita sa ma ajutati. Am ajuns la regula lui L'hopital. Dar nu inteleg cum se face astfel derivata
$\lim_{x \to \/2} \frac{\sqrt{x+2}-2}{x^2+6x+8}$

Answer 1

$\lim\limits_{x \to 2} \dfrac{\sqrt{x+2}-2}{x^2-6x+8} = \lim\limits_{x \to 2} \dfrac{\sqrt{x+2}-2}{(x-3)^2-1} =\lim\limits_{x \to 2} \dfrac{\sqrt{x+2}-2}{(x-3-1)(x-3+1)} = \\ \\ = \lim\limits_{x \to 2} \dfrac{\sqrt{x+2}-2}{(x-4)(x-2)} = \lim\limits_{x \to 2} \dfrac{\sqrt{x+2}-2}{(x-4)(x-2)} \overset{(*)}{=} \\ \\ \\ \sqrt{x+2} = t \Rightarrow x+2 = t^2\Rightarrow x = t^2-2 \\ x\rightarrow 2 \Rightarrow t \rightarrow \sqrt{2+2} \Rightarrow t\rightarrow \sqrt 4 \Rightarrow t \rightarrow 2$

$\overset{(*)}{=} \lim\limits_{t \to 2} \dfrac{t - 2}{(t^2-2-4)(t^2-2-2)} = \lim\limits_{t \to 2} \dfrac{t-2}{(t^2-6)(t^2-4)} = \\ \\ =\lim\limits_{t \to 2} \dfrac{t-2}{(t^2-6)(t-2)(t+2)} = \lim\limits_{t \to 2} \dfrac{1}{(t^2-6)(t+2)} = \dfrac{1}{(2^2-6)(2+2)} = \\ \\ = \dfrac{1}{(4-6)\cdot 4} = \dfrac{1}{-2\cdot 4} = \boxed{-\dfrac{1}{8}}$

Answer 2

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