Question Number 184903 by aba last updated on 13/Jan/23
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} −\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }=?? \\ $$$$\mathrm{let}\:\mathrm{x}=\mathrm{2t}\:\begin{cases}{\mathrm{x}\rightarrow\mathrm{0}}\\{\mathrm{t}\rightarrow\mathrm{0}}\end{cases} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} −\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{2t}} +\mathrm{e}^{−\mathrm{2t}} −\mathrm{2}}{\mathrm{4t}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{e}^{\mathrm{t}} −\mathrm{e}^{−\mathrm{t}} }{\mathrm{t}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{e}^{\mathrm{t}} −\mathrm{1}}{\mathrm{t}}+\frac{\mathrm{e}^{−\mathrm{t}} −\mathrm{1}}{−\mathrm{t}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\ $$
Answered by manxsol last updated on 13/Jan/23
$$ \\ $$$$\mathrm{1}+\frac{{x}}{\mathrm{1}!}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\mathrm{1}−\frac{{x}}{\mathrm{1}!}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!_{} }−\mathrm{2}=\frac{\mathrm{2}\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\mathrm{2}\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}}{{x}^{\mathrm{2}} }=\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{12}}……. \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{12}}+…..=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by aba last updated on 13/Jan/23
$$\mathrm{nice} \\ $$
Commented by manxsol last updated on 14/Jan/23
$$ \\ $$$$\frac{{e}^{\mathrm{0}.\mathrm{0000000001}} +{e}^{−\mathrm{0}.\mathrm{0000000001}} −\mathrm{2}}{\left(\mathrm{0}.\mathrm{0000000001}\right)^{\mathrm{2}} } \\ $$$$\mathrm{0}.\mathrm{0} \\ $$$${result}\:{of}\:{servidor} \\ $$$$\frac{{e}^{\mathrm{0}.\mathrm{00001}} +{e}^{−\mathrm{0}.\mathrm{00001}} −\mathrm{2}}{\mathrm{0}.\mathrm{00001}^{\mathrm{2}} } \\ $$$$\mathrm{1}.\mathrm{0} \\ $$$${result}\:{of}\:{servidor} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$