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Question Number 184903 by aba last updated on 13/Jan/23

lim_(x→0) ((e^x +e^(−x) −2)/x^2 )=?? let x=2t { ((x→0)),((t→0)) :} lim_(x→0) ((e^x +e^(−x) −2)/x^2 )=lim_(t→0) ((e^(2t) +e^(−2t) −2)/(4t^2 ))=(1/4)lim_(t→0) (((e^t −e^(−t) )/t))^2 =(1/4)lim_(t→0) (((e^t −1)/t)+((e^(−t) −1)/(−t)))^2 =(1/4)(1+1)^2 =1

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} −\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }=?? \\ $$$$\mathrm{let}\:\mathrm{x}=\mathrm{2t}\:\begin{cases}{\mathrm{x}\rightarrow\mathrm{0}}\\{\mathrm{t}\rightarrow\mathrm{0}}\end{cases} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} −\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{2t}} +\mathrm{e}^{−\mathrm{2t}} −\mathrm{2}}{\mathrm{4t}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{e}^{\mathrm{t}} −\mathrm{e}^{−\mathrm{t}} }{\mathrm{t}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{e}^{\mathrm{t}} −\mathrm{1}}{\mathrm{t}}+\frac{\mathrm{e}^{−\mathrm{t}} −\mathrm{1}}{−\mathrm{t}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\ $$

Answered by manxsol last updated on 13/Jan/23

1+(x/(1!))+(x^2 /(2!))+(x^3 /(3!))+1−(x/(1!))+(x^2 /(2!))−(x^3 /(3!_ ))−2=((2(x^2 /(2!))+2(x^4 /(4!)))/x^2 )=1+(x^2 /(12))....... lim_(x→0) 1+(x^2 /(12))+.....=1

$$ \\ $$$$\mathrm{1}+\frac{{x}}{\mathrm{1}!}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\mathrm{1}−\frac{{x}}{\mathrm{1}!}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!_{} }−\mathrm{2}=\frac{\mathrm{2}\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\mathrm{2}\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}}{{x}^{\mathrm{2}} }=\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{12}}....... \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{12}}+.....=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by aba last updated on 13/Jan/23

nice

$$\mathrm{nice} \\ $$

Commented by manxsol last updated on 14/Jan/23

((e^(0.0000000001) +e^(−0.0000000001) −2)/((0.0000000001)^2 )) 0.0 result of servidor ((e^(0.00001) +e^(−0.00001) −2)/(0.00001^2 )) 1.0 result of servidor

$$ \\ $$$$\frac{{e}^{\mathrm{0}.\mathrm{0000000001}} +{e}^{−\mathrm{0}.\mathrm{0000000001}} −\mathrm{2}}{\left(\mathrm{0}.\mathrm{0000000001}\right)^{\mathrm{2}} } \\ $$$$\mathrm{0}.\mathrm{0} \\ $$$${result}\:{of}\:{servidor} \\ $$$$\frac{{e}^{\mathrm{0}.\mathrm{00001}} +{e}^{−\mathrm{0}.\mathrm{00001}} −\mathrm{2}}{\mathrm{0}.\mathrm{00001}^{\mathrm{2}} } \\ $$$$\mathrm{1}.\mathrm{0} \\ $$$${result}\:{of}\:{servidor} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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