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Question Number 185537 by mathlove last updated on 23/Jan/23

lim_(x→0) ((e^(2x) −4e^x +2x+3)/x^3 )=?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{\mathrm{2}{x}} −\mathrm{4}{e}^{{x}} +\mathrm{2}{x}+\mathrm{3}}{{x}^{\mathrm{3}} }=? \\ $$

Commented by Ar Brandon last updated on 23/Jan/23

Same method as above e^(2x) →1+2x+2x^2 +((4x^3 )/3) e^x →1+x+(x^2 /2)+(x^3 /6)

$$\mathrm{Same}\:\mathrm{method}\:\mathrm{as}\:\mathrm{above} \\ $$$${e}^{\mathrm{2}{x}} \rightarrow\mathrm{1}+\mathrm{2}{x}+\mathrm{2}{x}^{\mathrm{2}} +\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{3}} \\ $$$${e}^{{x}} \rightarrow\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}} \\ $$

Answered by LEKOUMA last updated on 24/Jan/23

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

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

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