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Question Number 169711 by mnjuly1970 last updated on 06/May/22

prove that: lim_( x → 0) ( (1/x^( 2) ) − (e^( x) /((e^( x) −1 )^( 2) )) ) = (1/(12))

$$ \\ $$$$\:\:\:\:{prove}\:{that}: \\ $$$$\:\: \\ $$$$\:\:{lim}_{\:{x}\:\rightarrow\:\mathrm{0}} \left(\:\frac{\mathrm{1}}{{x}^{\:\mathrm{2}} }\:\:−\:\frac{{e}^{\:{x}} }{\left({e}^{\:{x}} −\mathrm{1}\:\right)^{\:\mathrm{2}} }\:\right)\:=\:\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\:\:\:\:\:\: \\ $$

Commented by infinityaction last updated on 06/May/22

i think question is wrong

$${i}\:{think}\:{question}\:{is}\:{wrong} \\ $$

Commented by cortano1 last updated on 07/May/22

no. the question right

$${no}.\:{the}\:{question}\:{right} \\ $$

Commented by infinityaction last updated on 07/May/22

thank you sir got my mistake

$${thank}\:{you}\:{sir}\: \\ $$$${got}\:{my}\:{mistake} \\ $$

Answered by qaz last updated on 07/May/22

L=lim_(x→0) ((1/x^2 )−(e^x /((e^x −1)^2 )))=lim_(x→0) ((e^(2x) −2e^x +1−x^2 e^x )/x^4 ) e^(2x) −2e^x +1−x^2 e^x =(1+2x+(1/2)(2x)^2 +(1/6)(2x)^3 +(1/(24))(2x)^4 +...)−2(1+x+(1/2)x^2 +(1/6)x^3 +(1/(24))x^4 +...)+ 1−x^2 (1+x+(1/2)x^2 +...) =(((16)/(24))−(2/(24))−(1/2))x^4 +...=(1/(12))x^4 +o(x^4 ) ⇒L=lim_(x→0) (((1/(12))x^4 +o(x^4 ))/x^4 )=(1/(12))

$$\mathrm{L}=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{e}^{\mathrm{x}} }{\left(\mathrm{e}^{\mathrm{x}} −\mathrm{1}\right)^{\mathrm{2}} }\right)=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{2x}} −\mathrm{2e}^{\mathrm{x}} +\mathrm{1}−\mathrm{x}^{\mathrm{2}} \mathrm{e}^{\mathrm{x}} }{\mathrm{x}^{\mathrm{4}} } \\ $$$$\mathrm{e}^{\mathrm{2x}} −\mathrm{2e}^{\mathrm{x}} +\mathrm{1}−\mathrm{x}^{\mathrm{2}} \mathrm{e}^{\mathrm{x}} \\ $$$$=\left(\mathrm{1}+\mathrm{2x}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2x}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{2x}\right)^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{24}}\left(\mathrm{2x}\right)^{\mathrm{4}} +...\right)−\mathrm{2}\left(\mathrm{1}+\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{6}}\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{24}}\mathrm{x}^{\mathrm{4}} +...\right)+ \\ $$$$\mathrm{1}−\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} +...\right) \\ $$$$=\left(\frac{\mathrm{16}}{\mathrm{24}}−\frac{\mathrm{2}}{\mathrm{24}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{x}^{\mathrm{4}} +...=\frac{\mathrm{1}}{\mathrm{12}}\mathrm{x}^{\mathrm{4}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{4}} \right) \\ $$$$\Rightarrow\mathrm{L}=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{\mathrm{12}}\mathrm{x}^{\mathrm{4}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{4}} \right)}{\mathrm{x}^{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{12}} \\ $$

Answered by cortano1 last updated on 07/May/22

lim_(x→0) (((e^x −1)^2 −x^2 e^x )/(x^2 (e^x −1)^2 )) = lim_(x→0) (((1+x+(x^2 /2)+(x^3 /6)+O(x^3 )−1)^2 −x^2 (1+x+(1/2)x^2 +(1/6)x^3 +O(x^3 )))/(x^2 (1+x+(1/2)x^2 +(1/6)x^3 +O(x^3 )−1)^2 )) = lim_(x→0) ((x+(1/2)x^2 +(1/6)x^3 +...−(x^2 +x^3 +(1/2)x^4 +(1/6)x^5 +...))/(x^2 (x+(1/2)x^2 +(1/6)x^3 +...)^2 )) = lim_(x→0) (((1/(12))x^4 +O(x^5 ))/(x^4 +O(x^5 ))) = (1/(12))

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left({e}^{{x}} −\mathrm{1}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} {e}^{{x}} }{{x}^{\mathrm{2}} \left({e}^{{x}} −\mathrm{1}\right)^{\mathrm{2}} }\: \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}+{O}\left({x}^{\mathrm{3}} \right)−\mathrm{1}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} \left(\mathrm{1}+{x}+\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} +{O}\left({x}^{\mathrm{3}} \right)\right)}{{x}^{\mathrm{2}} \left(\mathrm{1}+{x}+\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} +{O}\left({x}^{\mathrm{3}} \right)−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}+\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} +...−\left({x}^{\mathrm{2}} +{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{5}} +...\right)}{{x}^{\mathrm{2}} \left({x}+\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} +...\right)^{\mathrm{2}} } \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{\mathrm{12}}{x}^{\mathrm{4}} +{O}\left({x}^{\mathrm{5}} \right)}{{x}^{\mathrm{4}} +{O}\left({x}^{\mathrm{5}} \right)}\:=\:\frac{\mathrm{1}}{\mathrm{12}} \\ $$

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