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Question Number 51617 by Saorey last updated on 29/Dec/18

$$\mathrm{A}=\underset{x\to 0}{\lim }\frac{e^x-e^{-x}-2x}{x^3}(without$$$$using{L}^{\prime }\mathrm{hopital}\mathrm{rule})$$$$$$

Commented by aseerimad last updated on 29/Dec/18

how to do it by L'hopital ?

Commented by tanmay.chaudhury50@gmail.com last updated on 29/Dec/18

$$byLHrule$$$$\underset{x\to 0}{\lim }\frac{e^x-e^{-x}-2x}{x^3}\left(\frac{0}{0}\right)$$$$=\underset{x\to 0}{\lim }\frac{e^x+e^{-x}-2}{3x^2}\left(\frac{0}{0}\right)$$$$=\underset{x\to 0}{\lim }\frac{e^x-e^{-x}}{6x}\left(\frac{0}{0}\right)$$$$=\underset{x\to 0}{\lim }\frac{e^x+e^{-x}}{6}=\frac{2}{6}=\frac{1}{3}$$

Answered by Smail last updated on 29/Dec/18

$$e^x\underset{0}{\sim }1+x+\frac{x^2}{2!}+\frac{x^3}{3!}$$$$A=\underset{x\to 0}{lim}\frac{1+x+\frac{x^2}{2}+\frac{x^3}{6}-(1+(-x)+\frac{{(-x)}^2}{2}+\frac{{(-x)}^3}{6})-2x}{x^3}$$$$A=\underset{x\to 0}{lim}\frac{1+x+\frac{x^2}{2}+\frac{x^3}{6}-1+x-\frac{x^2}{2}+\frac{x^3}{6}-2x}{x^3}=\underset{x\to 0}{lim}\frac{2x+\frac{x^3}{3}-2x}{x^3}$$$$=\underset{x\to 0}{lim}\frac{\frac{x^3}{3}}{x^3}=\frac{1}{3}$$

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