Монгол Бодлогын Сан

Заавар:

Бодолт: $$\begin{align*} \text{Хяз.}&=\lim\limits_{x\to 0}\displaystyle\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt[3]{1+x}-\sqrt[3]{1-x}}\\ &=\lim\limits_{x\to 0}\dfrac{\color{red}{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}(\sqrt[3]{(1+x)^2}+\sqrt[3]{1+x}\cdot\sqrt[3]{1-x}+\sqrt[3]{(1-x)^2})}{\color{red}{(\sqrt[3]{1+x}-\sqrt[3]{1-x})(\sqrt[3]{(1+x)^2}+\sqrt[3]{1+x}\cdot\sqrt[3]{1-x}+\sqrt[3]{(1-x)^2})}(\sqrt{1+x}+\sqrt{1-x})}\\ &=\lim\limits_{x\to 0}\dfrac{\color{red}{\{(1+x)-(1-x)\}}(\sqrt[3]{(1+x)^2}+\sqrt[3]{1+x}\cdot\sqrt[3]{1-x}+\sqrt[3]{(1-x)^2})}{\color{red}{\{(1+x)-(1-x)\}}(\sqrt{1+x}+\sqrt{1-x})}\\ &=\lim\limits_{x\to 0}\dfrac{\sqrt[3]{(1+x)^2}+\sqrt[3]{1+x}\cdot\sqrt[3]{1-x}+\sqrt[3]{(1-x)^2}}{\sqrt{1+x}+\sqrt{1-x}}\\ &=\dfrac{\sqrt[3]{(1+0)^2}+\sqrt[3]{1+0}\cdot\sqrt[3]{1-0}+\sqrt[3]{(1-0)^2}}{\sqrt{1+0}+\sqrt{1-0}}=\dfrac{3}{2} \end{align*}$$