Монгол Бодлогын Сан

Заавар:

Бодолт: $$\begin{align*} \lim\limits_{x\to8}\dfrac{\sqrt{9+2x}-5}{\sqrt[3]{x}-2}&=\lim\limits_{x\to8}\dfrac{(\sqrt{9+2x}-5)(\sqrt{9+2x}+5)(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)}{(\sqrt[3]{x}-2)(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)(\sqrt{9+2x}+5)}\\ &=\lim\limits_{x\to8}\dfrac{((9+2x)-5^2)(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)}{(x-8)(\sqrt{9+2x}+5)}\\ &=\lim\limits_{x\to8}\dfrac{2\cancel{(x-8)}(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)}{\cancel{(x-8)}(\sqrt{9+2x}+5)}\\ &=\dfrac{2(\sqrt[3]{8^2}+2\sqrt[3]{8}+4)}{\sqrt{9+2\cdot 8}+5}=\dfrac{24}{10}=\dfrac{12}{5}\\ \end{align*}$$