Nhờ mn giúp mik bài này nha

Đáp án:

$L=\dfrac{11}{36}.$

Giải thích các bước giải:

$L=\displaystyle\lim_{x \to 0} \dfrac{\sqrt[3]{x+8}-2\sqrt{1+x}}{x^2-3x}\\ =\displaystyle\lim_{x \to 0} \dfrac{\sqrt[3]{x+8}-2-\sqrt{4x+4}+2}{x^2-3x}\\ =\displaystyle\lim_{x \to 0} \dfrac{\sqrt[3]{x+8}-2}{x^2-3x}-\displaystyle\lim_{x \to 0} \dfrac{\sqrt{4x+4}-2}{x^2-3x}\\ =\displaystyle\lim_{x \to 0} \dfrac{(\sqrt[3]{x+8}-2)(\sqrt[3]{x+8}^2+2\sqrt[3]{x+8}+4)}{x(x-3)(\sqrt[3]{x+8}^2+2\sqrt[3]{x+8}+4)}-\displaystyle\lim_{x \to 0} \dfrac{(\sqrt{4x+4}-2)(\sqrt{4x+4}+2)}{x(x-3)(\sqrt{4x+4}+2)}\\ =\displaystyle\lim_{x \to 0} \dfrac{x+8-8}{x(x-3)(\sqrt[3]{x+8}^2+2\sqrt[3]{x+8}+4)}-\displaystyle\lim_{x \to 0} \dfrac{4x+4-4}{x(x-3)(\sqrt{4x+4}+2)}\\ =\displaystyle\lim_{x \to 0} \dfrac{x}{x(x-3)(\sqrt[3]{x+8}^2+2\sqrt[3]{x+8}+4)}-\displaystyle\lim_{x \to 0} \dfrac{4x}{x(x-3)(\sqrt{4x+4}+2)}\\ =\displaystyle\lim_{x \to 0} \dfrac{1}{(x-3)(\sqrt[3]{x+8}^2+2\sqrt[3]{x+8}+4)}-\displaystyle\lim_{x \to 0} \dfrac{4}{(x-3)(\sqrt{4x+4}+2)}\\ =\dfrac{1}{(0-3)(\sqrt[3]{0+8}^2+2\sqrt[3]{0+8}+4)}-\dfrac{4}{(0-3)(\sqrt{4.0+4}+2)}\\ =\dfrac{11}{36}.$