Question

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Answer 1

$1)\; \; \lim\limits _{x \to 0}\frac{2x}{\sqrt{4+x}-\sqrt{4-x}}=\lim\limits _{x \to 0}\frac{2x(\sqrt{4+x}+\sqrt{4-x})}{4+x-(4-x)}=\lim\limits_{x \to 0}(\sqrt{4+x}+\sqrt{4-x})=\\\\=\sqrt{4+0}+\sqrt{4-0}=2+2=4\\\\2)\; \; \lim\limits _{x \to 0}\frac{2-\sqrt{x}}{3-\sqrt{2x+1}}=\frac{2-0}{3-\sqrt1}=\frac{2}{2}=1$

$3)\; \; \lim\limits _{x \to 8}\frac{\sqrt[3]{x}-2}{\sqrt{x}-2\sqrt2}=\lim\limits _{x \to 8}\frac{\sqrt[3]{x}-2}{\sqrt{x}-\sqrt8}=\lim\limits _{x \to 8}\frac{(\sqrt[3]{x}-2)(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)(\sqrt{x}+\sqrt8)}{(\sqrt{x}-\sqrt8)(\sqrt{x}+\sqrt8)(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)}=\\\\=\lim\limits_{x \to 8}\frac{(x-8)(\sqrt{x}+\sqrt8)}{(x-8)(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)}=\frac{\sqrt8+\sqrt8}{\sqrt[3]{8^2}+2\sqrt[3]8+4}=\frac{4\sqrt2}{4+4+4}=\frac{\sqrt2}{3}$