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Question Number 78865 by jagoll last updated on 21/Jan/20

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

Commented by jagoll last updated on 21/Jan/20

$$\mathrm{dear}\mathrm{mr}\mathrm{Mjs}.\mathrm{what}\mathrm{value}\mathrm{the}\mathrm{limit}?$$$${\mathrm{i}}^{\prime }\mathrm{m}\mathrm{got}\frac{1}{24}.\mathrm{my}\mathrm{answer}\mathrm{is}\mathrm{correct}?$$

Commented by john santu last updated on 21/Jan/20

$$\sqrt[3]{1+\sqrt{1+x^3}}-\sqrt[3]{2}=$$$$\frac{1+\sqrt{1+x^3}-2}{\sqrt[3]{(1+\sqrt{{1+x^3)}^2}}+\sqrt[3]{2(1+\sqrt{1+x^3)}}+\sqrt[3]{4}}\times \frac{1}{2x^3}=$$$$\underset{x\to 0}{\lim }\frac{\sqrt{1+x^3}-1}{2x^3}\times \frac{1}{3\sqrt[3]{4}}=$$$$\frac{1}{3\sqrt[3]{4}}\underset{x\to 0}{\lim }\frac{(1+\frac{1}{2}{x}^3)-1}{2x^3}=$$$$\frac{1}{3\sqrt[3]{4}}\times \frac{1}{4}=\frac{1}{12\sqrt[3]{4}}\times \frac{\sqrt[3]{2}}{\sqrt[3]{2}}=\frac{\sqrt[3]{2}}{24}.$$

Commented by john santu last updated on 21/Jan/20

$$wrongsir$$

Commented by jagoll last updated on 21/Jan/20

$$\frac{\sqrt[3]{2}}{24}\mathrm{sir}$$

Commented by jagoll last updated on 21/Jan/20

$$\mathrm{yes}\mathrm{sir}.\mathrm{i}\mathrm{agree}.$$

Answered by MJS last updated on 21/Jan/20

$$\frac{\frac{d}{dx}[\sqrt[3]{1+\sqrt{1+x^3}}-\sqrt[3]{2}]}{\frac{d}{dx}\left[2x^3\right]}=\frac{\frac{x^2}{2\sqrt{1+x^3}\sqrt[3]{{(1+\sqrt{1+x^3})}^2}}}{6x^2}=$$$$=\frac{1}{12\sqrt{1+x^3}\sqrt[3]{{(1+\sqrt{1+x^3})}^2}}$$$$\mathrm{and}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{this}\mathrm{with}x=0\mathrm{is}\frac{\sqrt[3]{2}}{24}$$

Commented by jagoll last updated on 21/Jan/20

$$\mathrm{thanks}\mathrm{sir}$$

Commented by Henri Boucatchou last updated on 21/Jan/20

$$\mathit{A}lrightwithHospitalRule$$

Answered by behi83417@gmail.com last updated on 22/Jan/20

$$1+{\mathrm{x}}^3={\mathrm{t}}^2[\mathrm{x}\to 0\Rightarrow \mathrm{t}\to 1]$$$$\mathrm{L}=\underset{\mathrm{t}\to 1}{\lim }\frac{\sqrt[3]{1+\mathrm{t}}-\sqrt[3]{2}}{2({\mathrm{t}}^2-1)}$$$$1+\mathrm{t}={\mathrm{r}}^3[\mathrm{t}\to 1\Rightarrow \mathrm{r}\to \sqrt[3]{2}]$$$$\mathrm{L}=\underset{\mathrm{r}\to \sqrt[3]{2}}{\lim }\frac{\mathrm{r}-\sqrt[3]{2}}{2[{({\mathrm{r}}^3-1)}^2-1]}=\underset{\mathrm{r}\to \sqrt[3]{2}}{\lim }\frac{1}{2\left[{6\mathrm{r}}^2({\mathrm{r}}^3-1)\right]}=$$$$=\frac{1}{2\times 6\sqrt[3]{4}}=\frac{\sqrt[3]{2}}{24}$$

Commented by john santu last updated on 22/Jan/20

$$wrongsir.theequation\sqrt{1+x^3}$$$$not1+x^3$$

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