Question and Answers Forum

All Questions      Topic List

Vector Questions

Previous in All Question      Next in All Question      

Previous in Vector      Next in Vector      

Question Number 171033 by cortano1 last updated on 06/Jun/22

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

Commented by greougoury555 last updated on 07/Jun/22

$$\sqrt[3]{\frac{(1-\sqrt{1-x})(1+\sqrt{1-x})}{1+\sqrt{1-x}}.\frac{\sqrt{1+x}+1}{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}}$$$$=\sqrt[3]{\frac{x(\sqrt{1+x}+1)}{x(1+\sqrt{1-x})}}=\sqrt[3]{\frac{\sqrt{1+x}+1}{1+\sqrt{1-x}}}$$$$L=\underset{x\to 0}{\lim }\frac{1}{x}\left(\frac{\sqrt[3]{\sqrt{1+x}+1}-\sqrt[3]{1+\sqrt{1-x}}}{\sqrt[3]{1+\sqrt{1-x}}}\right)$$$$=\frac{1}{\sqrt[3]{2}}.\underset{x\to 0}{\lim }\frac{\sqrt[3]{\sqrt{1+x}+1}-\sqrt[3]{1+\sqrt{1-x}}}{x}$$$$=\frac{1}{\sqrt[3]{2}}.\underset{x\to 0}{\lim }\frac{(\sqrt{1+x}+1)-(1+\sqrt{1-x})}{x[\sqrt[3]{{(\sqrt{1+x}+1)}^2}+\sqrt[3]{(\sqrt{1+x}+1)(1+\sqrt{1-x})}+\sqrt[3]{{(1+\sqrt{1-x})}^2}}$$$$=\frac{1}{3.\sqrt[3]{8}}.\underset{x\to 0}{\lim }\frac{\sqrt{1+x}-\sqrt{1-x}}{x}$$$$=\frac{1}{6}.\underset{x\to 0}{\lim }\frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})}$$$$=\frac{1}{6}.\frac{2}{2}=\frac{1}{6}$$

Answered by qaz last updated on 06/Jun/22

$$\underset{\mathrm{x}\to 0}{\lim }\frac{1}{\mathrm{x}}[\sqrt[3]{\frac{1-\sqrt{1-\mathrm{x}}}{\sqrt{1+\mathrm{x}}-1}}-1]$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{1}{\mathrm{x}}\ln \sqrt[3]{\frac{1-\sqrt{1-\mathrm{x}}}{\sqrt{1+\mathrm{x}}-1}}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{1}{3\mathrm{x}}\ln \frac{1-\sqrt{1-\mathrm{x}}}{\sqrt{1+\mathrm{x}}-1}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{1}{3\mathrm{x}}\cdot (\frac{1-\sqrt{1-\mathrm{x}}}{\sqrt{1+\mathrm{x}}-1}-1)$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{2-\sqrt{1-\mathrm{x}}-\sqrt{1+\mathrm{x}}}{3\mathrm{x}(\sqrt{1+\mathrm{x}}-1)}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{\frac{1}{4}{\mathrm{x}}^2+\mathrm{o}\left({\mathrm{x}}^2\right)}{\frac{3}{2}{\mathrm{x}}^2+\mathrm{o}\left({\mathrm{x}}^2\right)}$$$$=\frac{1}{6}$$

Answered by qaz last updated on 06/Jun/22

$$\underset{\mathrm{x}\to 0}{\lim }\frac{1}{\mathrm{x}}[\sqrt[3]{\frac{1-\sqrt{1-\mathrm{x}}}{\sqrt{1+\mathrm{x}}-1}}-1]$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{1}{\mathrm{x}}[\sqrt[3]{\frac{\frac{1}{2}\mathrm{x}+\frac{1}{8}{\mathrm{x}}^2+\mathrm{o}\left({\mathrm{x}}^2\right)}{\frac{1}{2}\mathrm{x}-\frac{1}{8}{\mathrm{x}}^2+\mathrm{o}\left({\mathrm{x}}^2\right)}}-1]$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{1}{\mathrm{x}}[\sqrt[3]{\frac{1+\frac{1}{4}\mathrm{x}+\mathrm{o}\left(\mathrm{x}\right)}{1-\frac{1}{4}\mathrm{x}+\mathrm{o}\left(\mathrm{x}\right)}}-1]$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{1}{\mathrm{x}}[\sqrt[3]{(1+\frac{1}{4}\mathrm{x}+\mathrm{o}\left(\mathrm{x}\right))(1+\frac{1}{4}\mathrm{x}+\mathrm{o}\left(\mathrm{x}\right))}-1]$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{1}{\mathrm{x}}[\sqrt[3]{1+\frac{1}{2}\mathrm{x}+\mathrm{o}\left(\mathrm{x}\right)}-1]$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{1}{\mathrm{x}}(\frac{1}{6}\mathrm{x}+\mathrm{o}\left(\mathrm{x}\right))$$$$=\frac{1}{6}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com