Question Number 87153 by jagoll last updated on 03/Apr/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} −\mathrm{2}}{\mathrm{x}^{\mathrm{2}} } \\ $$
Commented by john santu last updated on 03/Apr/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{−\mathrm{x}} }{\mathrm{2x}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} }{\mathrm{2}}\:=\:\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 03/Apr/20
$${we}\:{have}\:{e}^{{x}} \:=\mathrm{1}+\frac{{x}}{\mathrm{1}!}\:+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}\:+{o}\left({x}^{\mathrm{3}} \right)\:\:\left({x}\rightarrow\mathrm{0}\right) \\ $$$${e}^{−{x}} =\mathrm{1}−\frac{{x}}{\mathrm{1}!}\:+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}\:+{o}\left({x}^{\mathrm{3}} \right)\:\Rightarrow{e}^{{x}} \:+{e}^{−{x}} −\mathrm{2}\:=\:{x}^{\mathrm{2}} \:+{o}\left({x}^{\mathrm{3}} \right)\:\Rightarrow \\ $$$$\frac{{e}^{{x}} \:+{e}^{−{x}} −\mathrm{1}}{{x}^{\mathrm{2}} }\:=\mathrm{1}+{o}\left({x}\right)\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{e}^{{x}} \:+{e}^{−{x}} −\mathrm{2}}{{x}^{\mathrm{2}} }=\mathrm{1} \\ $$
Answered by $@ty@m123 last updated on 03/Apr/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{e}^{\mathrm{x}} +\frac{\mathrm{1}}{{e}^{{x}} }−\mathrm{2}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{e}^{\mathrm{2}{x}} +\mathrm{1}−\mathrm{2}{e}^{{x}} }{\mathrm{x}^{\mathrm{2}} .{e}^{{x}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left({e}^{{x}} −\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} .{e}^{{x}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{{e}^{{x}} −\mathrm{1}}{{x}}\right)^{\mathrm{2}} ×\frac{\mathrm{1}}{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:{e}^{{x}} } \\ $$$$=\mathrm{1}^{\mathrm{2}} ×\mathrm{1} \\ $$$$=\mathrm{1} \\ $$