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Question Number 113651 by bemath last updated on 14/Sep/20

$$\underset{x\to 0}{\lim }(\frac{1}{2}-\frac{1}{1+e^{-x}}).\frac{1}{3x}=?$$

Answered by john santu last updated on 14/Sep/20

$$byTaylorseries$$$$letf\left(x\right)=\frac{1}{1+e^{-x}}=\frac{1}{2}+\frac{x}{4}-\frac{x^3}{48}+\frac{x^5}{480}-...$$$$then\underset{x\to 0}{\lim }\frac{\frac{1}{2}-\frac{1}{1+e^{-x}}}{3x}=$$$$\underset{x\to 0}{\lim }\frac{\frac{1}{2}-(\frac{1}{2}+\frac{x}{4}-\frac{x^3}{48}+...)}{3x}=$$$$\underset{x\to 0}{\lim }\frac{-\frac{x}{4}+\frac{x^3}{48}-...}{3x}=-\frac{1}{12}$$

Answered by bobhans last updated on 14/Sep/20

$$\underset{x\to 0}{\lim }\left(\frac{1+{\mathrm{e}}^{-\mathrm{x}}-2}{6\mathrm{x}(1+{\mathrm{e}}^{-\mathrm{x}})}\right)=\underset{x\to 0}{\lim }\left(\frac{{\mathrm{e}}^{-\mathrm{x}}-1}{6\mathrm{x}(1+{\mathrm{e}}^{-\mathrm{x}})}\right)$$$$\underset{x\to 0}{\lim }\left(\frac{-{\mathrm{e}}^{-\mathrm{x}}}{6+{6\mathrm{e}}^{-\mathrm{x}}-{6\mathrm{xe}}^{-\mathrm{x}}}\right)=-\frac{1}{12}$$$$\mathrm{via}{\mathrm{L}}^{\prime }\mathrm{Hopital}\mathrm{rule}$$

Answered by Dwaipayan Shikari last updated on 14/Sep/20

$$(\frac{1}{2}-\frac{e^x}{1+e^x})\frac{1}{3x}=\frac{1+e^x-2e^x}{2(1+e^x)}.\frac{1}{3x}=\frac{1}{6}\left(\frac{1-e^x}{2.x}\right)=\frac{1}{12}\left(\frac{e^x-1}{-x}\right)=-\frac{1}{12}$$

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