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L-lim-x-0-e-x-cos-x-x-1-x-3-

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Question Number 141900 by iloveisrael last updated on 24/May/21
L = lim_(x→0) ((e^x cos x−x−1)/x^3 ) =?
$$\:\mathcal{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} \:\mathrm{cos}\:{x}−{x}−\mathrm{1}}{{x}^{\mathrm{3}} }\:=? \\ $$
Answered by Dwaipayan Shikari last updated on 24/May/21
lim_(x→0) ((e^x (cosx)−x−1)/x^3 )=((e^x (1−(x^2 /2))−x−1)/x^3 ) =((e^x −x−1−e^x (x^2 /2))/x^3 )=(((x^2 /2)+(x^3 /(3!))+O(x^4 )−(1+x+O(x^2 ))(x^2 /2))/x^3 ) O(x^n )→0 ⇒(((x^3 /(3!))−(x^3 /3))/x^3 )=−(1/6)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}} \left({cosx}\right)−{x}−\mathrm{1}}{{x}^{\mathrm{3}} }=\frac{{e}^{{x}} \left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)−{x}−\mathrm{1}}{{x}^{\mathrm{3}} } \\ $$$$=\frac{{e}^{{x}} −{x}−\mathrm{1}−{e}^{{x}} \frac{{x}^{\mathrm{2}} }{\mathrm{2}}}{{x}^{\mathrm{3}} }=\frac{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+{O}\left({x}^{\mathrm{4}} \right)−\left(\mathrm{1}+{x}+{O}\left({x}^{\mathrm{2}} \right)\right)\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}{{x}^{\mathrm{3}} } \\ $$$${O}\left({x}^{{n}} \right)\rightarrow\mathrm{0}\:\:\:\:\Rightarrow\frac{\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}}{{x}^{\mathrm{3}} }=−\frac{\mathrm{1}}{\mathrm{6}} \\ $$
Answered by mathmax by abdo last updated on 25/May/21
f(x)=((e^x cosx−x−1)/x^3 ) we have e^x ∼1+x+(x^2 /2) and cosx∼1−(x^2 /2) ⇒ e^x cosx∼(1+x+(x^2 /2))(1−(x^2 /2))=1−(x^2 /2) +x−(x^3 /2)+(x^2 /2)−(x^4 /4) =1+x−(x^3 /2)−(x^4 /4) ⇒f(x)∼((1+x−(x^3 /2)−(x^4 /4)−x−1)/x^3 )=−(1/2)−(x/4) ⇒ lim_(x→0) f(x)=−(1/2)
$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{e}^{\mathrm{x}} \mathrm{cosx}−\mathrm{x}−\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{e}^{\mathrm{x}} \:\sim\mathrm{1}+\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:\:\mathrm{and}\:\mathrm{cosx}\sim\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{e}^{\mathrm{x}} \:\mathrm{cosx}\sim\left(\mathrm{1}+\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right)=\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:+\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}} \\ $$$$=\mathrm{1}+\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{2}}−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\mathrm{1}+\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{2}}−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}}−\mathrm{x}−\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }=−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{x}}{\mathrm{4}}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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