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lim-x-0-x-e-x-1-2-e-x-1-x-3-




Question Number 153039 by ZiYangLee last updated on 04/Sep/21
lim_(x→0)  ((x(e^x +1)−2(e^x −1))/x^3 ) = ?
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left({e}^{{x}} +\mathrm{1}\right)−\mathrm{2}\left({e}^{{x}} −\mathrm{1}\right)}{{x}^{\mathrm{3}} }\:=\:? \\ $$
Answered by puissant last updated on 04/Sep/21
=lim_(x→0) ((x(1+x+(x^2 /(2!))+x^2 ε(x)+1)−2(1+x+(x^2 /(2!))+(x^3 /6)+x^3 ε(x)−1))/x^3 )  =lim_(x→0) ((2x+x^2 +(x^3 /2)−2x−x^2 −(x^3 /3)+x^3 ε(x))/x^3 )  =lim_(x→0) ((x^3 ((1/2)−(1/3)+ε(x)))/x^3 )  ; x→0 ⇒ ε(x)→0.  =  (1/2) − (1/3)   =  (1/6).
$$={lim}_{{x}\rightarrow\mathrm{0}} \frac{{x}\left(\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+{x}^{\mathrm{2}} \varepsilon\left({x}\right)+\mathrm{1}\right)−\mathrm{2}\left(\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}+{x}^{\mathrm{3}} \varepsilon\left({x}\right)−\mathrm{1}\right)}{{x}^{\mathrm{3}} } \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{2}{x}+{x}^{\mathrm{2}} +\frac{{x}^{\mathrm{3}} }{\mathrm{2}}−\mathrm{2}{x}−{x}^{\mathrm{2}} −\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+{x}^{\mathrm{3}} \varepsilon\left({x}\right)}{{x}^{\mathrm{3}} } \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \frac{{x}^{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}+\varepsilon\left({x}\right)\right)}{{x}^{\mathrm{3}} }\:\:;\:{x}\rightarrow\mathrm{0}\:\Rightarrow\:\varepsilon\left({x}\right)\rightarrow\mathrm{0}. \\ $$$$=\:\:\frac{\mathrm{1}}{\mathrm{2}}\:−\:\frac{\mathrm{1}}{\mathrm{3}}\:\:\:=\:\:\frac{\mathrm{1}}{\mathrm{6}}. \\ $$

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