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Question Number 195126 by Erico last updated on 25/Jul/23

Soit f_n (x)=2^(n+1) [(((1/2^n )cotan((x/2^n ))−cotanx)/(sin((x/2^n ))))] Calculer lim_(x→0) f_n (x) et lim_(n→+∞) ((f_n (x))/2^(2n+2) )

$$\mathrm{Soit}\:{f}_{{n}} \left({x}\right)=\mathrm{2}^{{n}+\mathrm{1}} \left[\frac{\frac{\mathrm{1}}{\mathrm{2}^{{n}} }{cotan}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)−{cotanx}}{{sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)}\right] \\ $$$${Calculer}\:\underset{{x}\rightarrow\mathrm{0}} {{lim}f}_{{n}} \left({x}\right)\:{et}\:\underset{{n}\rightarrow+\infty} {{lim}}\:\frac{{f}_{{n}} \left({x}\right)}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{2}} } \\ $$

Answered by MM42 last updated on 25/Jul/23

f_n (x)=((2sinxcos((x/2^n ))−2^(n+1) sin((x/2^n ))cosx)/(sin^2 ((x/2^n ))sinx)) =((sin(x+(x/2^n ))+sin(x−(x/2^n ))−2^n sin((x/2^n )+x)−2^n sin((x/2^n )−x))/(sin^2 ((x/2^n ))sinx)) =(((1−2^n )sin(((1+2^n )/2^n ))x−(1+2^n )sin(((1−2^n )/2^n )))/(sin^2 ((x/2^n ))sinx)) =2^n ×((((1−2^n )/2^n )sin(((1+2^n )/2^n ))x−((1+2^n )/2^n )sin(((1−2^n )/2^n )))/(sin^2 ((x/2^n ))sinx)) Tip :if x→0 ⇒ asinbx−bsinax ∼ ((ab(a−b)(a+b))/6)x^3 ⇒lim_(x→0) f_n (x)=lim_(x→0) 2^n ×(((((1−2^n )/2^n ))(((1+2^n )/2^n ))(((1−2^n )/2^n )−((1+2^n )/2^n ))(((1−2^n )/2^n )+((1+2^n )/2^n ))(x^3 /6))/(((x/2^n ))^2 ×x)) = (((1−2^(2n) )(−2^(n+1) )×2^(3n) (2))/(2^(4n) ×6))= =((2^(2n+1) −2)/3) ✓ lim_(n→∞) ((f_n (x))/2^(2n+2) ) =(1/6) ✓

$${f}_{{n}} \left({x}\right)=\frac{\mathrm{2}{sinxcos}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)−\mathrm{2}^{{n}+\mathrm{1}} {sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right){cosx}}{{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}^{{n}} }\right){sinx}} \\ $$$$=\frac{{sin}\left({x}+\frac{{x}}{\mathrm{2}^{{n}} }\right)+{sin}\left({x}−\frac{{x}}{\mathrm{2}^{{n}} }\right)−\mathrm{2}^{{n}} {sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }+{x}\right)−\mathrm{2}^{{n}} {sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }−{x}\right)}{{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}^{{n}} }\right){sinx}} \\ $$$$=\frac{\left(\mathrm{1}−\mathrm{2}^{{n}} \right){sin}\left(\frac{\mathrm{1}+\mathrm{2}^{{n}} }{\mathrm{2}^{{n}} }\right){x}−\left(\mathrm{1}+\mathrm{2}^{{n}} \right){sin}\left(\frac{\mathrm{1}−\mathrm{2}^{{n}} }{\mathrm{2}^{{n}} }\right)}{{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}^{{n}} }\right){sinx}} \\ $$$$=\mathrm{2}^{{n}} ×\frac{\frac{\mathrm{1}−\mathrm{2}^{{n}} }{\mathrm{2}^{{n}} }{sin}\left(\frac{\mathrm{1}+\mathrm{2}^{{n}} }{\mathrm{2}^{{n}} }\right){x}−\frac{\mathrm{1}+\mathrm{2}^{{n}} }{\mathrm{2}^{{n}} }{sin}\left(\frac{\mathrm{1}−\mathrm{2}^{{n}} }{\mathrm{2}^{{n}} }\right)}{{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}^{{n}} }\right){sinx}} \\ $$$${Tip}\::{if}\:\:{x}\rightarrow\mathrm{0}\:\Rightarrow\:{asinbx}−{bsinax}\:\sim\:\frac{{ab}\left({a}−{b}\right)\left({a}+{b}\right)}{\mathrm{6}}{x}^{\mathrm{3}} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:{f}_{{n}} \left({x}\right)={lim}_{{x}\rightarrow\mathrm{0}} \:\mathrm{2}^{{n}} ×\frac{\left(\frac{\mathrm{1}−\mathrm{2}^{{n}} }{\mathrm{2}^{{n}} }\right)\left(\frac{\mathrm{1}+\mathrm{2}^{{n}} }{\mathrm{2}^{{n}} }\right)\left(\frac{\mathrm{1}−\mathrm{2}^{{n}} }{\mathrm{2}^{{n}} }−\frac{\mathrm{1}+\mathrm{2}^{{n}} }{\mathrm{2}^{{n}} }\right)\left(\frac{\mathrm{1}−\mathrm{2}^{{n}} }{\mathrm{2}^{{n}} }+\frac{\mathrm{1}+\mathrm{2}^{{n}} }{\mathrm{2}^{{n}} }\right)\frac{{x}^{\mathrm{3}} }{\mathrm{6}}}{\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)^{\mathrm{2}} ×{x}} \\ $$$$=\:\frac{\left(\mathrm{1}−\mathrm{2}^{\mathrm{2}{n}} \right)\left(−\mathrm{2}^{{n}+\mathrm{1}} \right)×\mathrm{2}^{\mathrm{3}{n}} \left(\mathrm{2}\right)}{\mathrm{2}^{\mathrm{4}{n}} ×\mathrm{6}}= \\ $$$$=\frac{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} −\mathrm{2}}{\mathrm{3}}\:\:\checkmark \\ $$$$ \\ $$$${lim}_{{n}\rightarrow\infty} \:\frac{{f}_{{n}} \left({x}\right)}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{6}}\:\checkmark \\ $$$$ \\ $$

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