Question Number 213128 by klipto last updated on 30/Oct/24
$$\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}^{+\:} } \frac{\mathrm{2}}{\mathrm{1}+\boldsymbol{\mathrm{e}}^{−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}} } \\ $$
Answered by a.lgnaoui last updated on 30/Oct/24
$$\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}^{+\:} } \frac{\mathrm{2}}{\mathrm{1}+\boldsymbol{\mathrm{e}}^{−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}} } \\ $$$$\:\:\:\frac{\mathrm{2}}{\mathrm{1}+\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{x}}} }=\frac{\mathrm{2}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}}} }}\:\:=\:\:\frac{\mathrm{2}}{\frac{\mathrm{1}+\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}}} }{\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}}} }}=\:\frac{\mathrm{2e}^{\frac{\mathrm{1}}{\mathrm{x}}} }{\mathrm{1}+\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}}} } \\ $$$$\mathrm{donc}\:\mathrm{lim}\:_{\mathrm{x}\rightarrow\mathrm{o}^{+} } \frac{\mathrm{2}}{\left(\mathrm{1}+\mathrm{e}^{\frac{−\mathrm{1}}{\mathrm{x}}} \right)}=\mathrm{2}\: \\ $$$$ \\ $$
Answered by MrGaster last updated on 30/Oct/24
$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{2}}{\mathrm{1}+{e}^{−\frac{\mathrm{1}}{{x}}} }=\frac{\mathrm{2}}{\mathrm{1}+{e}^{−\infty} }=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{0}}=\frac{\mathrm{2}}{\mathrm{1}}=\mathrm{2} \\ $$