Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

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)

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{e}^{\mathrm{x}} \mathrm{cosx}−\mathrm{x}−\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{e}^{\mathrm{x}} \:\sim\mathrm{1}+\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:\:\mathrm{and}\:\mathrm{cosx}\sim\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{e}^{\mathrm{x}} \:\mathrm{cosx}\sim\left(\mathrm{1}+\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right)=\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:+\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}} \\ $$$$=\mathrm{1}+\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{2}}−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\mathrm{1}+\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{2}}−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}}−\mathrm{x}−\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }=−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{x}}{\mathrm{4}}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com