Question Number 173380 by cortano1 last updated on 10/Jul/22
Commented by kaivan.ahmadi last updated on 10/Jul/22
$$\sim{li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\:\:\frac{{e}^{{x}} −{e}^{−{x}} +\mathrm{2}{sinx}}{\mathrm{1}}=\mathrm{0} \\ $$$$ \\ $$
Answered by CElcedricjunior last updated on 12/Jul/22
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\:\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} −\mathrm{1}+\boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{x}}} −\mathrm{1}−\mathrm{2}\left(\boldsymbol{\mathrm{cosx}}−\mathrm{1}\right)}{\boldsymbol{\mathrm{x}}} \\ $$$$\:\:\:\:\:\:=\mathrm{0} \\ $$$$\: \\ $$$$\:....{le}\:{c}\acute {{e}l}\grave {{e}bre}\:{c}\acute {{e}dric}\:{junior}....... \\ $$