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Question Number 162643 by tounghoungko last updated on 31/Dec/21

$$\underset{x\to \mathrm{\infty }}{\lim }{x}^{4/3}(\sqrt[3]{x^2+1}+\sqrt[3]{3-x^2})=?$$

Answered by Ar Brandon last updated on 31/Dec/21

$$\mathcal{L}=\underset{x\to \mathrm{\infty }}{\lim }{x}^{4/3}(\sqrt[3]{x^2+1}+\sqrt[3]{3-x^2})$$$$=\underset{x\to \mathrm{\infty }}{\lim }{x}^{4/3}\cdot x^{\frac{2}{3}}(1+\frac{1}{3x^2}-(1-\frac{3}{3x^2}))$$$$=\underset{x\to \mathrm{\infty }}{\lim }{x}^2\left(\frac{4}{3x^2}\right)=\frac{4}{3}$$

Commented by Ar Brandon last updated on 31/Dec/21

$$\mathrm{Can}\mathrm{someone}\mathrm{please}\mathrm{check}.\mathrm{I}\mathrm{have}\mathrm{some}\mathrm{doubt}.$$

Commented by MJS_new last updated on 31/Dec/21

$$\mathrm{strange}\mathrm{idea},\mathrm{not}\mathrm{sure}\mathrm{if}\mathrm{it}\mathrm{makes}\mathrm{any}\mathrm{sense}$$$$\left(\mathrm{but}\mathrm{anyway}\mathrm{your}\mathrm{result}\mathrm{is}\mathrm{right}\right)$$$$x^{4/3}({(x^2+1)}^{1/3}-{(x^2-3)}^{1/3})=\mathcal{L}$$$$a^{1/3}-b^{1/3}=c$$$$a-b-3a^{1/3}{b}^{1/3}(a^{1/3}-b^{1/3})=c^3$$$$3a^{1/3}{b}^{1/3}c=a-b-c^3$$$$27{abc}^3={(a-b-c^3)}^3$$$$\mathrm{inserting}\mathrm{\&}\mathrm{transforming}$$$$x^{12}-\frac{{\mathcal{L}}^3(54x^{10}-33x^8-12{\mathcal{L}}^3{x}^4+{\mathcal{L}}^6)}{27{\mathcal{L}}^3-64}=0$$$$\Rightarrow \mathcal{L}\ne \frac{4}{3}$$$$\mathrm{which}\mathrm{means}\mathrm{we}\mathrm{cannot}\mathrm{reach}\mathcal{L}=\frac{3}{4}\mathrm{while}x\in \mathbb{R}$$

Commented by Ar Brandon last updated on 31/Dec/21

$$\mathrm{Thanks}\mathrm{for}\mathrm{confirming},\mathrm{Sir}.$$

Answered by tounghoungko last updated on 31/Dec/21

$$\underset{x\to \mathrm{\infty }}{\lim }{x}^{4/3}(x^{2/3}\sqrt[3]{1+\frac{1}{x^2}}-x^{2/3}\sqrt[3]{-\frac{3}{x^2}+1})$$$$[\frac{1}{x^2}=y]$$$$=\underset{y\to 0}{\lim }\frac{\sqrt[3]{1+y}-\sqrt[3]{1-3y}}{y}=\underset{y\to 0}{\lim }\frac{(1+\frac{y}{3})-(1-\frac{3y}{3})}{y}$$$$=\underset{y\to 0}{\lim }\frac{\frac{4}{3}y}{y}=\frac{4}{3}$$

Commented by tounghoungko last updated on 31/Dec/21

$$\underset{x\to \mathrm{\infty }}{\lim }{x}^{4/3}(\sqrt[3]{x^2+1}+\sqrt[3]{3-x^2})$$$$=\underset{x\to \mathrm{\infty }}{\lim }{x}^{4/3}(\sqrt[3]{x^2+1}+\sqrt[3]{-(x^2-3)})$$$$=\underset{x\to \mathrm{\infty }}{\lim }{x}^{4/3}(\sqrt[3]{x^2+1}-\sqrt[3]{x^2-3})$$$$=\underset{x\to 0}{\lim }{x}^2(\sqrt[3]{1+\frac{1}{x^2}}-\sqrt[3]{1-\frac{3}{x^2}})$$$$=\underset{y\to 0}{\lim }\frac{\sqrt[3]{1+y}-\sqrt[3]{1-3y}}{y}$$$$=\underset{y\to 0}{\lim }\frac{(1+\frac{y}{3})-(1-\frac{3y}{3})}{y}$$$$=\underset{y\to 0}{\lim }\frac{\frac{4}{3}y}{y}=\frac{4}{3}$$

Answered by mathmax by abdo last updated on 31/Dec/21

$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{\frac{4}{3}}\{{(1+{\mathrm{x}}^2)}^{\frac{1}{3}}-\left({}^3\sqrt{3}\right){(1-\frac{{\mathrm{x}}^2}{3})}^{\frac{1}{3}}\}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{\frac{4}{3}}\{{\mathrm{x}}^{\frac{2}{3}}{(1+\frac{1}{{\mathrm{x}}^2})}^{\frac{1}{3}}-\left({}^3\sqrt{3}\right){\mathrm{x}}^{\frac{2}{3}}{(\frac{1}{{\mathrm{x}}^2}-\frac{1}{3})}^{\frac{1}{3}}\}$$$$={\mathrm{x}}^2\{{(1+\frac{1}{{\mathrm{x}}^2})}^{\frac{1}{3}}-{(1-\frac{3}{{\mathrm{x}}^2})}^{\frac{1}{3}}\}$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)\sim {\mathrm{x}}^2(1+\frac{1}{{3\mathrm{x}}^2}-(1-\frac{1}{{\mathrm{x}}^2}))={\mathrm{x}}^2(\frac{1}{{3\mathrm{x}}^2}+\frac{3}{{3\mathrm{x}}^2})=\frac{4}{3}\Rightarrow$$$${\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)=\frac{4}{3}$$

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