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Question Number 145009 by mim24 last updated on 01/Jul/21

Answered by ArielVyny last updated on 01/Jul/21

lim_(x→0) ((xcosx)/(sinx))+((xcos(2x))/(sinx)) lim_(x→0) (x/(tgx))+(x/(sinx))(2sin^2 x+1)=(x/(tgx))+2xsinx+(x/(sinx)) lim(2xsinx)=0(x→0) lim_(x→0) (x/(tgx))=lim_(x→0) (1/(1+tg^2 x))=1 (hospital theorem′s) lim_(x→0) (x/(sinx))=lim_(x→0) (1/(cosx))=1 finally lim_(x→0) ((x(cosx+cos2x))/(sinx))=2

$${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{xcosx}}{{sinx}}+\frac{{xcos}\left(\mathrm{2}{x}\right)}{{sinx}} \\ $$$${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{x}}{{tgx}}+\frac{{x}}{{sinx}}\left(\mathrm{2}{sin}^{\mathrm{2}} {x}+\mathrm{1}\right)=\frac{{x}}{{tgx}}+\mathrm{2}{xsinx}+\frac{{x}}{{sinx}} \\ $$$${lim}\left(\mathrm{2}{xsinx}\right)=\mathrm{0}\left({x}\rightarrow\mathrm{0}\right) \\ $$$${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{x}}{{tgx}}={li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{\mathrm{1}}{\mathrm{1}+{tg}^{\mathrm{2}} {x}}=\mathrm{1}\:\:\left({hospital}\:{theorem}'{s}\right) \\ $$$${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{x}}{{sinx}}={li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{\mathrm{1}}{{cosx}}=\mathrm{1} \\ $$$${finally}\:{li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{x}\left({cosx}+{cos}\mathrm{2}{x}\right)}{{sinx}}=\mathrm{2} \\ $$

Answered by Rasheed.Sindhi last updated on 01/Jul/21

lim_(x→0) ((x(cosx+cos2x))/(sinx)) =lim_(x→0) ((cosx+cos2x)/((sinx)/x)) =((lim_(x→0) (cosx+cos2x))/(lim_(x→0) (((sinx)/x)))) =((cos0+cos(2×0))/1)=1+1=2

$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{x}\left({cosx}+{cos}\mathrm{2}{x}\right)}{{sinx}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{cosx}+{cos}\mathrm{2}{x}}{\frac{{sinx}}{{x}}} \\ $$$$=\frac{\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left({cosx}+{cos}\mathrm{2}{x}\right)}{\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\frac{{sinx}}{{x}}\right)} \\ $$$$=\frac{{cos}\mathrm{0}+{cos}\left(\mathrm{2}×\mathrm{0}\right)}{\mathrm{1}}=\mathrm{1}+\mathrm{1}=\mathrm{2} \\ $$

Commented by greg_ed last updated on 01/Jul/21

cool, sir Rasheed !

$$\mathrm{cool},\:\mathrm{sir}\:\mathrm{Rasheed}\:! \\ $$

Commented by Rasheed.Sindhi last updated on 01/Jul/21

Th𝛂nX sir greg_ed!

$$\boldsymbol{\mathcal{T}{h}\alpha{n}\mathcal{X}}\:\:\mathrm{sir}\:\mathrm{greg\_ed}! \\ $$

Answered by mathmax by abdo last updated on 01/Jul/21

we know lim_(x→0) ((sinx)/x)=1 ⇒lim_(x→0) (x/(sinx))=1 ⇒ lim_(x⌣0) ((x(cosx +cos(2x)))/(sinx))=lim_(x→0) (x/(sinx))×lim_(x→0) (cosx +cos(2x)) =1×(1+1)=2

$$\mathrm{we}\:\mathrm{know}\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{sinx}}{\mathrm{x}}=\mathrm{1}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \frac{\mathrm{x}}{\mathrm{sinx}}=\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\smile\mathrm{0}} \frac{\mathrm{x}\left(\mathrm{cosx}\:+\mathrm{cos}\left(\mathrm{2x}\right)\right)}{\mathrm{sinx}}=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \frac{\mathrm{x}}{\mathrm{sinx}}×\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \left(\mathrm{cosx}\:+\mathrm{cos}\left(\mathrm{2x}\right)\right) \\ $$$$=\mathrm{1}×\left(\mathrm{1}+\mathrm{1}\right)=\mathrm{2} \\ $$

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