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Question Number 142351 by bramlexs22 last updated on 30/May/21

$$\underset{x\to 0}{\lim }\frac{\sqrt[3]{27+x}-\sqrt[3]{27-x}}{\sqrt[3]{x^2}+\sqrt[4]{x^3}}=?$$

Answered by mathmax by abdo last updated on 30/May/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{{(27+\mathrm{x})}^{\frac{1}{3}}-{(27-\mathrm{x})}^{\frac{1}{3}}}{{\mathrm{x}}^{\frac{2}{3}}+{\mathrm{x}}^{\frac{3}{4}}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{3{(1+\frac{\mathrm{x}}{27})}^{\frac{1}{3}}-3{(1-\frac{\mathrm{x}}{27})}^{\frac{1}{3}}}{{\mathrm{x}}^{\frac{2}{3}}+{\mathrm{x}}^{\frac{3}{4}}}$$$$\sim \frac{3.(1+\frac{1}{3.27}\mathrm{x})-3(1-\frac{\mathrm{x}}{3.27})}{{\mathrm{x}}^{\frac{2}{3}}+{\mathrm{x}}^{\frac{3}{4}}}=\frac{\frac{2}{27}\mathrm{x}}{{\mathrm{x}}^{\frac{2}{3}}(1+{\mathrm{x}}^{\frac{3}{4}-\frac{2}{3}})}$$$$=\frac{1}{27}{\mathrm{x}}^{\frac{1}{3}}\times \frac{1}{1+{\mathrm{x}}^{\frac{1}{12}}}=\frac{\left({}^3\sqrt{\mathrm{x}}\right)}{27(1+\left({}^{12}\sqrt{\mathrm{x}}\right)}\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)=0$$

Commented by mathmax by abdo last updated on 30/May/21

$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{2\left({}^3\sqrt{\mathrm{x}}\right)}{27(1+\left({}^{12}\sqrt{\mathrm{x}}\right)}$$

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