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Question Number 147459 by liberty last updated on 21/Jul/21

$$\underset{x\to 0}{\lim }\frac{\sqrt{1+x^2}\sqrt[3]{8+x^3}-2}{x^2}=?$$

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

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{1+{\mathrm{x}}^2}.\left({}^3\sqrt{8+{\mathrm{x}}^3}\right)-2}{{\mathrm{x}}^2}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{1+{\mathrm{x}}^2}.\left(2{(1+\frac{{\mathrm{x}}^3}{8})}^{\frac{1}{3}}\right)-2}{{\mathrm{x}}^2}$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)\sim \frac{2(1+\frac{{\mathrm{x}}^2}{2})(1+\frac{{\mathrm{x}}^3}{24})-2}{{\mathrm{x}}^2}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{2(1+\frac{{\mathrm{x}}^3}{24}+\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^5}{48})-2}{{\mathrm{x}}^2}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)\sim \frac{\frac{{\mathrm{x}}^3}{12}+{\mathrm{x}}^2+\frac{{\mathrm{x}}^5}{24}}{{\mathrm{x}}^2}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)\sim 1+\frac{\mathrm{x}}{12}+\frac{{\mathrm{x}}^3}{24}\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)=1$$

Answered by gsk2684 last updated on 21/Jul/21

$$=\underset{x\to 0}{\lim }\frac{{(1+x^2)}^{\frac{1}{2}}2{(1+\frac{x^3}{8})}^{\frac{1}{3}}-2}{x^2}$$$$=2\underset{x\to 0}{\lim }\frac{(1+\frac{x^2}{2}+\frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}{x}^4+...)(1+\frac{1}{3}\frac{x^3}{8}+...)-1}{x^2}$$$$=2\underset{x\to 0}{\lim }\frac{(1+\frac{x^2}{2})\left(1\right)-1}{x^2}$$$$=1$$

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