$\lim_{n \to -3} \frac{\sqrt[3]{x+1}-\sqrt{3-x}}{x}$ 21/08/2021 Bởi Julia $\lim_{n \to -3} \frac{\sqrt[3]{x+1}-\sqrt{3-x}}{x}$
$\lim\limits_{x\to -3}\dfrac{\sqrt[3]{x+1}-\sqrt{3-x}}{x}$ $=\dfrac{\sqrt[3]{-3+1}-\sqrt{3+3}}{-3}$ $=\dfrac{\sqrt[3]{2}+\sqrt6}{3}$ Bình luận
$\displaystyle\lim_{n \to -3} \dfrac{\sqrt[3]{x+1}-\sqrt{3-x}}{x}\\ =\dfrac{\sqrt[3]{-3+1}-\sqrt{3+3}}{-3}\\ =\dfrac{-\sqrt[3]{2}-\sqrt{6}}{-3}\\ =\dfrac{\sqrt[3]{2}+\sqrt{6}}{3}$ Bình luận
$\lim\limits_{x\to -3}\dfrac{\sqrt[3]{x+1}-\sqrt{3-x}}{x}$
$=\dfrac{\sqrt[3]{-3+1}-\sqrt{3+3}}{-3}$
$=\dfrac{\sqrt[3]{2}+\sqrt6}{3}$
$\displaystyle\lim_{n \to -3} \dfrac{\sqrt[3]{x+1}-\sqrt{3-x}}{x}\\ =\dfrac{\sqrt[3]{-3+1}-\sqrt{3+3}}{-3}\\ =\dfrac{-\sqrt[3]{2}-\sqrt{6}}{-3}\\ =\dfrac{\sqrt[3]{2}+\sqrt{6}}{3}$