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Question Number 213128 by klipto last updated on 30/Oct/24

$${\mathbf{lim}}_{\mathbf{x}\to 0^{+}}\frac{2}{1+{\mathbf{e}}^{-\frac{1}{\mathbf{x}}}}$$

Answered by a.lgnaoui last updated on 30/Oct/24

$${\mathbf{lim}}_{\mathbf{x}\to 0^{+}}\frac{2}{1+{\mathbf{e}}^{-\frac{1}{\mathbf{x}}}}$$$$\frac{2}{1+{\mathrm{e}}^{-\frac{1}{\mathrm{x}}}}=\frac{2}{1+\frac{1}{{\mathrm{e}}^{\frac{1}{\mathrm{x}}}}}=\frac{2}{\frac{1+{\mathrm{e}}^{\frac{1}{\mathrm{x}}}}{{\mathrm{e}}^{\frac{1}{\mathrm{x}}}}}=\frac{{2\mathrm{e}}^{\frac{1}{\mathrm{x}}}}{1+{\mathrm{e}}^{\frac{1}{\mathrm{x}}}}$$$$\mathrm{donc}\lim {}_{\mathrm{x}\to {\mathrm{o}}^{+}}\frac{2}{(1+{\mathrm{e}}^{\frac{-1}{\mathrm{x}}})}=2$$$$$$

Answered by MrGaster last updated on 30/Oct/24

$$\underset{x\to 0^{+}}{\lim }\frac{2}{1+e^{-\frac{1}{x}}}=\frac{2}{1+e^{-\mathrm{\infty }}}=\frac{2}{1+0}=\frac{2}{1}=2$$

Answered by Frix last updated on 31/Oct/24

$$\frac{2}{1+{\mathrm{e}}^{-\frac{1}{x}}}=1+\tanh \frac{1}{2x}$$$$\underset{x\to 0^{+}}{\lim }\frac{2}{1+{\mathrm{e}}^{-\frac{1}{x}}}=\underset{x\to 0^{+}}{\lim }(1+\tanh \frac{1}{2x})=$$$$=1+\underset{t\to +\mathrm{\infty }}{\lim }\tanh t=1+1=2$$

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