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Question Number 140905 by bramlexs22 last updated on 14/May/21

lim_(x→0) ((1−cos (1−cos x))/(x (tan x−x))) =?

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

Answered by bramlexs22 last updated on 14/May/21

Answered by mathmax by abdo last updated on 14/May/21

let L(x)=((1−cos(1−cosx))/(x(tanx−x))) we have 1−cosx ∼(x^2 /2) ⇒cos(1−cosx)∼cos((x^2 /2))∼1−(x^4 /8) ⇒ 1−cos(1−cosx)∼(x^4 /8) tanx ∼x+(x^3 /3) ⇒tanx −x ∼(x^3 /3) ⇒x(tanx−x)∼(x^4 /3) ⇒ f(x)∼(x^4 /8).(3/x^4 ) ⇒lim_(x→0) f(x)=(3/8)

$$\mathrm{let}\:\mathrm{L}\left(\mathrm{x}\right)=\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{1}−\mathrm{cosx}\right)}{\mathrm{x}\left(\mathrm{tanx}−\mathrm{x}\right)} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{1}−\mathrm{cosx}\:\sim\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\mathrm{cos}\left(\mathrm{1}−\mathrm{cosx}\right)\sim\mathrm{cos}\left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right)\sim\mathrm{1}−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{8}}\:\Rightarrow \\ $$$$\mathrm{1}−\mathrm{cos}\left(\mathrm{1}−\mathrm{cosx}\right)\sim\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{8}} \\ $$$$\mathrm{tanx}\:\sim\mathrm{x}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}\:\Rightarrow\mathrm{tanx}\:−\mathrm{x}\:\sim\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}\:\Rightarrow\mathrm{x}\left(\mathrm{tanx}−\mathrm{x}\right)\sim\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{3}}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{8}}.\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{4}} }\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{3}}{\mathrm{8}} \\ $$

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