Tính các giới hạn sau: $\lim\limits_{x \to 2}$$\dfrac{\sqrt{x^2+x+10}-2\sqrt{x^2-2x+4}}{x^2-4}$

Đáp án:

$-\dfrac{3}{32}.$

Giải thích các bước giải:

$\displaystyle\lim_{x \to 2} \dfrac{\sqrt{x^2+x+10}-2\sqrt{x^2-2x+4}}{x^2-4}\\ =\displaystyle\lim_{x \to 2} \dfrac{\sqrt{x^2+x+10}-2\sqrt{2}.\sqrt{x}-2\sqrt{x^2-2x+4}+2\sqrt{2}.\sqrt{x}}{x^2-4}\\ =\displaystyle\lim_{x \to 2} \dfrac{\sqrt{x^2+x+10}-2\sqrt{2}.\sqrt{x}}{x^2-4}-\displaystyle\lim_{x \to 2} \dfrac{2\sqrt{x^2-2x+4}-2\sqrt{2}.\sqrt{x}}{x^2-4}\\ =\displaystyle\lim_{x \to 2} \dfrac{\left(\sqrt{x^2+x+10}-2\sqrt{2}.\sqrt{x} \right)\left(\sqrt{x^2+x+10}+2\sqrt{2}.\sqrt{x} \right)}{(x^2-4)\left(\sqrt{x^2+x+10}+2\sqrt{2}.\sqrt{x} \right)}-\displaystyle\lim_{x \to 2} \dfrac{\left(2\sqrt{x^2-2x+4}-2\sqrt{2}.\sqrt{x} \right)\left(2\sqrt{x^2-2x+4}+2\sqrt{2}.\sqrt{x} \right)}{(x^2-4)\left(2\sqrt{x^2-2x+4}+2\sqrt{2}.\sqrt{x} \right)}\\ =\displaystyle\lim_{x \to 2} \dfrac{x^2-7x+10}{(x-2)(x+2)\left(\sqrt{x^2+x+10}+2\sqrt{2}.\sqrt{x} \right)}-4\displaystyle\lim_{x \to 2} \dfrac{x^2-4x+4}{(x-2)(x+2)\left(2\sqrt{x^2-2x+4}+2\sqrt{2}.\sqrt{x} \right)}\\ =\displaystyle\lim_{x \to 2} \dfrac{(x-2)(x-5)}{(x-2)(x+2)\left(\sqrt{x^2+x+10}+2\sqrt{2}.\sqrt{x} \right)}-4\displaystyle\lim_{x \to 2} \dfrac{(x-2)^2}{(x-2)(x+2)\left(2\sqrt{x^2-2x+4}+2\sqrt{2}.\sqrt{x} \right)}\\ =\displaystyle\lim_{x \to 2} \dfrac{x-5}{(x+2)\left(\sqrt{x^2+x+10}+2\sqrt{2}.\sqrt{x} \right)}-4\displaystyle\lim_{x \to 2} \dfrac{x-2}{(x+2)\left(2\sqrt{x^2-2x+4}+2\sqrt{2}.\sqrt{x} \right)}\\ = \dfrac{2-5}{(2+2)\left(\sqrt{2^2+2+10}+2\sqrt{2}.\sqrt{2}\right)}-4.\dfrac{2-2}{(2+2)\left(2\sqrt{x^2-2.2+4}+2\sqrt{2}.\sqrt{2} \right)}\\ =-\dfrac{3}{32}.$