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Question Number 148439 by liberty last updated on 28/Jul/21

$$\underset{x\to 2}{\lim }\frac{\sqrt{\mathrm{x}+2}\sqrt[3]{\mathrm{x}+6}-{\mathrm{x}}^2}{\mathrm{x}-2}=?$$

Answered by dumitrel last updated on 28/Jul/21

$$\underset{x\to 2}{lim}\frac{\sqrt{x+2}(\sqrt[3]{x+6}-2)}{x-2}+\underset{x\to 2}{lim}\frac{2(\sqrt{x+2}-2)}{x-2}-\underset{x\to 2}{lim}\frac{x^2-4}{x-2}=$$$$\underset{x\to 2}{lim}\frac{\sqrt{x+2}(x-2)}{(x-2)(\sqrt[3]{{(x+6)}^2}+2\sqrt[3]{x+2}+4}+\underset{x\to 2}{lim}\frac{2(x-2)}{(x-2)(\sqrt{x+2}+2)}+\underset{x\to 2}{lim}\frac{(x-2)(x+2)}{x-2)}=$$$$=\frac{2}{4+4+4}+\frac{2}{2+2}-4=-\frac{10}{3}$$

Answered by EDWIN88 last updated on 28/Jul/21

Answered by mathmax by abdo last updated on 28/Jul/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{x}+2}{(\mathrm{x}+6)}^{\frac{1}{3}}-{\mathrm{x}}^2}{\mathrm{x}-2}\mathrm{changement}\mathrm{x}-2=\mathrm{t}\mathrm{give}$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{t}\right)=\frac{\sqrt{\mathrm{t}+4}{(\mathrm{t}+8)}^{\frac{1}{3}}-{(\mathrm{t}+2)}^2}{\mathrm{t}}$$$$=\frac{\sqrt{\mathrm{t}+4}{(\mathrm{t}+8)}^{\frac{1}{3}}-{\mathrm{t}}^2-4\mathrm{t}-4}{\mathrm{t}}=\frac{2\sqrt{1+\frac{\mathrm{t}}{4}}({}^3\sqrt{8}{(1+\frac{\mathrm{t}}{8})}^{\frac{1}{3}}-{\mathrm{t}}^2-4\mathrm{t}-4}{\mathrm{t}}$$$$\Rightarrow \mathrm{g}\left(\mathrm{t}\right)\sim \frac{4(1+\frac{\mathrm{t}}{8})(1+\frac{\mathrm{t}}{24})-{\mathrm{t}}^2-4\mathrm{t}-4}{\mathrm{t}}$$$$=\frac{4(1+\frac{\mathrm{t}}{24}+\frac{\mathrm{t}}{8}+\frac{{\mathrm{t}}^2}{8.24})-{\mathrm{t}}^2-4\mathrm{t}-4}{\mathrm{t}}$$$$=\frac{\frac{\mathrm{t}}{6}+\frac{\mathrm{t}}{2}+\frac{{\mathrm{t}}^2}{48}-{\mathrm{t}}^2-4\mathrm{t}}{\mathrm{t}}=\frac{1}{6}+\frac{1}{2}-4+(\frac{\mathrm{t}}{48}-\mathrm{t})\Rightarrow$$$${\lim }_{\mathrm{t}\to 0}\mathrm{g}\left(\mathrm{t}\right)=\frac{2}{3}-4=-\frac{10}{3}\Rightarrow {\lim }_{\mathrm{x}⌣2}\mathrm{f}\left(\mathrm{x}\right)=-\frac{10}{3}$$

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