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Question Number 124915 by liberty last updated on 06/Dec/20

$$\underset{x\to 0}{\lim }\frac{2\tan x-5\sin x+3x-x^3}{1-\sqrt[5]{1-2x^3}}=?$$

Answered by bemath last updated on 07/Dec/20

$$\underset{x\to 0}{\lim }\frac{2(x+\frac{x^3}{3})-5(x-\frac{x^3}{6})+3x-x^3}{1-(1-\frac{2x^3}{5})}=$$$$\underset{x\to 0}{\lim }\frac{\frac{2x^3}{3}+\frac{5x^3}{6}-x^3}{\frac{2x^3}{5}}=\underset{x\to 0}{\lim }\frac{\frac{1}{2}{x}^3}{\frac{2x^3}{5}}=\frac{5}{4}$$

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