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 144556 by imjagoll last updated on 26/Jun/21

lim_(x→0) (((1−cos x(√(cos 2x)))/x^2 )) =?

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{x}\sqrt{\mathrm{cos}\:\mathrm{2x}}}{\mathrm{x}^{\mathrm{2}} }\right)\:=? \\ $$

Answered by liberty last updated on 26/Jun/21

lim_(x→0) (((1−cos x(√(cos 2x)))/x^2 ))=? Sol: lim_(x→0) (((1−cos x+cos x−cos x(√(cos 2x)))/x^2 )) = lim_(x→0) (((1−cos x)/x^2 ))+lim_(x→0) (((cos x−cos x(√(cos 2x)))/x^2 )) = 2lim_(x→0) (((sin (1/2)x)/x))^2 +lim_(x→0) (((cos x(1−(√(cos 2x))))/x^2 )) =2.(1/4)+lim_(x→0) (((cos x(1−cos 2x))/((1+(√(cos 2x)))x^2 ))) =(1/2)+lim_(x→0) (((cos x)/( 1+(√(cos 2x))))).lim_(x→0) (((1−cos 2x)/x^2 )) =(1/2)+(1/2).lim_(x→0) (((2sin^2 x)/x^2 )) =(1/2)+1=(3/2). ▼

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{x}\sqrt{\mathrm{cos}\:\mathrm{2x}}}{\mathrm{x}^{\mathrm{2}} }\right)=? \\ $$$$\mathrm{Sol}:\: \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}\sqrt{\mathrm{cos}\:\mathrm{2x}}}{\mathrm{x}^{\mathrm{2}} }\right) \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{x}}{\mathrm{x}^{\mathrm{2}} }\right)+\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{cos}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}\sqrt{\mathrm{cos}\:\mathrm{2x}}}{\mathrm{x}^{\mathrm{2}} }\right) \\ $$$$=\:\mathrm{2}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}}{\mathrm{x}}\right)^{\mathrm{2}} +\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{cos}\:\mathrm{x}\left(\mathrm{1}−\sqrt{\mathrm{cos}\:\mathrm{2x}}\right)}{\mathrm{x}^{\mathrm{2}} }\right) \\ $$$$=\mathrm{2}.\frac{\mathrm{1}}{\mathrm{4}}+\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{cos}\:\mathrm{x}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2x}\right)}{\left(\mathrm{1}+\sqrt{\mathrm{cos}\:\mathrm{2x}}\right)\mathrm{x}^{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}+\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{cos}\:\mathrm{x}}{\:\mathrm{1}+\sqrt{\mathrm{cos}\:\mathrm{2x}}}\right).\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2x}}{\mathrm{x}^{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2sin}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{x}^{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}=\frac{\mathrm{3}}{\mathrm{2}}.\:\blacktrinagledown \\ $$

Answered by mathmax by abdo last updated on 26/Jun/21

f(x)=((1−cosx(√(cos(2x))))/x^2 ) we have cosx∼1−(x^2 /2) cos(2x)∼1−2x^2 ⇒(√(cos(2x)))∼(√(1−2x^2 ))∼1−x^2 ⇒ f(x)∼((1−(1−(x^2 /2))(1−x^2 ))/x^2 )=((1−(1−x^2 −(x^2 /2)+(x^4 /2)))/x^2 ) =(((3/2)x^2 −(x^4 /2))/x^2 )=(3/2)−(x^2 /2) ⇒lim_(x→0) f(x)=(3/2)

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com