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Question Number 82950 by jagoll last updated on 26/Feb/20

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

Commented by mathmax by abdo last updated on 26/Feb/20

$$wehave{a}^3-b^3=(a-b)(a^2+ab+b^2)\Rightarrow$$$$a-b=\left({}^3\sqrt{a}{-}^3\sqrt{b}\right)({\left({}^3\sqrt{a}\right)}^2{+}^3\sqrt{ab}+{\left({}^3\sqrt{b}\right)}^2)\Rightarrow$$$${\Rightarrow }^3\frac{1}{x^3}\left(\sqrt{1+tanx}{-}^3\sqrt{1+sinx}\right)=\frac{1}{x^3}\times \left(\frac{1+tanx-1-sinx}{{(1+tanx)}^{\frac{2}{3}}+(1+tanx){(1+sinx)}^{\frac{1}{3}}+{(1+sinx)}^{\frac{2}{3}}}\right)\Rightarrow$$$${lim}_{x\to 0}(...)={lim}_{x\to 0}\frac{1}{3x^3}(\frac{sinx}{cosx}-sinx)=\frac{1}{3}{lim}_{x\to 0}\frac{sinx}{x}\times \frac{1-cosx}{x^2\times cosx}$$$$=\frac{1}{3}\times 1\times \frac{1}{2}=\frac{1}{6}$$

Answered by john santu last updated on 26/Feb/20

$$\underset{x\to 0}{\lim }\frac{\tan \mathrm{x}(1-\cos \mathrm{x})}{{\mathrm{x}}^3}\times \underset{x\to 0}{\lim }\frac{1}{\sqrt[3]{{(1+\tan \mathrm{x})}^2}+\sqrt[3]{{(1+\sin \mathrm{x})}^2}+\sqrt[3]{(1+\tan \mathrm{x})(1+\sin \mathrm{x})}}$$$$=\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}$$

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