Question Number 146994 by KONE last updated on 17/Jul/21
$${u}_{{n}} ={cos}\sqrt{{n}+\mathrm{1}}−{cos}\sqrt{{n}} \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{u}_{{n}} =?? \\ $$
Answered by mathmax by abdo last updated on 17/Jul/21
$$\mathrm{u}_{\mathrm{n}} =\mathrm{cos}\left(\sqrt{\mathrm{n}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}\right)}−\mathrm{cos}\left(\sqrt{\mathrm{n}}\right)\right. \\ $$$$=\mathrm{cos}\left(\sqrt{\mathrm{n}}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}}\right)−\mathrm{cos}\left(\sqrt{\mathrm{n}}\right) \\ $$$$\sim\mathrm{cos}\left(\sqrt{\mathrm{n}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2n}}\right)\right)−\mathrm{cos}\left(\sqrt{\mathrm{n}}\right) \\ $$$$=\mathrm{cos}\left(\sqrt{\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{n}}}\right)−\mathrm{cos}\left(\sqrt{\mathrm{n}}\right) \\ $$$$\mathrm{cosp}−\mathrm{cosq}=\mathrm{cosp}+\mathrm{cos}\left(\pi−\mathrm{q}\right)=\mathrm{2cos}\left(\frac{\mathrm{p}+\pi−\mathrm{q}}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\mathrm{p}−\pi+\mathrm{q}}{\mathrm{2}}\right) \\ $$$$=\mathrm{2cos}\left(\frac{\pi}{\mathrm{2}}+\frac{\mathrm{p}−\mathrm{q}}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\mathrm{p}+\mathrm{q}}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}\right)=−\mathrm{2sin}\left(\frac{\mathrm{p}−\mathrm{q}}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\mathrm{p}+\mathrm{q}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$$\mathrm{u}_{\mathrm{n}} \sim−\mathrm{2sin}\left(\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{n}}}\right)\mathrm{cos}\left(\sqrt{\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{n}}}\right)\:\Rightarrow \\ $$$$\mid\mathrm{u}_{\mathrm{n}} \mid\leqslant\mathrm{2}\mid\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{n}}}\right)\mid\:\:\:\mathrm{due}\:\mathrm{to}\:\mid\mathrm{cos}\left(\sqrt{\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{n}}}\right)\mid\leqslant\mathrm{1}\:\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{n}}}\right)=\mathrm{0}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{u}_{\mathrm{n}} =\mathrm{0} \\ $$
Answered by mathmax by abdo last updated on 17/Jul/21
,(√(n+1))[ tel que u_n =((√(n+1))−(√n)) (−(1/(2(√c)))sin(√c)) (f(x)=cos((√x))) ⇒u_n =−(((√(n+1))−(√n))/(2(√c)))sin(√c) =−(1/(2(√c)((√(n+1))+(√n))))sin(√c) ⇒ ∣u_n ∣≤(1/(2(√c)((√(n+1))+(√n))))→0 (n→+∞) ⇒lim_(n→+∞) u_n =0](https://www.tinkutara.com/question/Q146999.png)
$$\mathrm{autre}\:\mathrm{method}\:\:\mathrm{on}\:\mathrm{utilise}\:\mathrm{le}\:\mathrm{theoreme}\:\mathrm{des}\:\mathrm{accroissements}\:\mathrm{finis}\:\Rightarrow \\ $$$$\left.\Rightarrow\exists\xi\:\in\right]\sqrt{\mathrm{n}},\sqrt{\mathrm{n}+\mathrm{1}}\left[\:\:\mathrm{tel}\:\mathrm{que}\:\mathrm{u}_{\mathrm{n}} =\left(\sqrt{\mathrm{n}+\mathrm{1}}−\sqrt{\mathrm{n}}\right)\:\left(−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{c}}}\mathrm{sin}\sqrt{\mathrm{c}}\right)\:\:\left(\mathrm{f}\left(\mathrm{x}\right)=\mathrm{cos}\left(\sqrt{\mathrm{x}}\right)\right)\right. \\ $$$$\Rightarrow\mathrm{u}_{\mathrm{n}} =−\frac{\sqrt{\mathrm{n}+\mathrm{1}}−\sqrt{\mathrm{n}}}{\mathrm{2}\sqrt{\mathrm{c}}}\mathrm{sin}\sqrt{\mathrm{c}}\:\:=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{c}}\left(\sqrt{\mathrm{n}+\mathrm{1}}+\sqrt{\mathrm{n}}\right)}\mathrm{sin}\sqrt{\mathrm{c}}\:\Rightarrow \\ $$$$\mid\mathrm{u}_{\mathrm{n}} \mid\leqslant\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{c}}\left(\sqrt{\mathrm{n}+\mathrm{1}}+\sqrt{\mathrm{n}}\right)}\rightarrow\mathrm{0}\:\left(\mathrm{n}\rightarrow+\infty\right)\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{u}_{\mathrm{n}} =\mathrm{0} \\ $$
Answered by mindispower last updated on 17/Jul/21
![f(x)=cos((√x)),mean value th ⇒∀(a,b) ∃_c ∈]a,b[ such ⇒f(a)−f(b)=f′(c)(a−b) ⇒cos((√(n+1)))−cos((√n))=−(1/(2(√c)))sin((√c))(n+1−n) ⇒∣cos((√(n+1)))−cos((√n))∣=(1/(2(√c_n ))) ∣sin(c_n )∣ usinc c_n ∈]n,n+1⌊,∣sin∣<1 ⇒∣cos((√(n+1)))−cos((√n))∣≤(1/(2(√n))) 2 nd way cos(a)−cos(b)=− 2sin(((a−b)/2))sin(((a+b)/2)) a=(√(n+1)),b=(√n) a−b=(1/( (√(n+1))+(√n))) and using sa m ,sin(a)<1 ⇔∣cos((√(n+1)))−cos((√n))∣≤2sin((1/( (√(n+1))+(√n))))](https://www.tinkutara.com/question/Q147029.png)
$${f}\left({x}\right)={cos}\left(\sqrt{{x}}\right),{mean}\:{value}\:{th} \\ $$$$\left.\Rightarrow\forall\left({a},{b}\right)\:\exists_{{c}} \in\right]{a},{b}\left[\:{such}\right. \\ $$$$\Rightarrow{f}\left({a}\right)−{f}\left({b}\right)={f}'\left({c}\right)\left({a}−{b}\right) \\ $$$$\Rightarrow{cos}\left(\sqrt{{n}+\mathrm{1}}\right)−{cos}\left(\sqrt{{n}}\right)=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{{c}}}{sin}\left(\sqrt{{c}}\right)\left({n}+\mathrm{1}−{n}\right) \\ $$$$\Rightarrow\mid{cos}\left(\sqrt{\left.{n}+\mathrm{1}\right)}−{cos}\left(\sqrt{{n}}\right)\mid=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{c}_{{n}} }}\:\:\:\mid{sin}\left({c}_{{n}} \right)\mid\right. \\ $$$$\left.{usinc}\:{c}_{{n}} \in\right]{n},{n}+\mathrm{1}\lfloor,\mid{sin}\mid<\mathrm{1} \\ $$$$\Rightarrow\mid{cos}\left(\sqrt{{n}+\mathrm{1}}\right)−{cos}\left(\sqrt{{n}}\right)\mid\leqslant\frac{\mathrm{1}}{\mathrm{2}\sqrt{{n}}} \\ $$$$\mathrm{2}\:{nd}\:{way} \\ $$$${cos}\left({a}\right)−{cos}\left({b}\right)=−\:\mathrm{2}{sin}\left(\frac{{a}−{b}}{\mathrm{2}}\right){sin}\left(\frac{{a}+{b}}{\mathrm{2}}\right) \\ $$$${a}=\sqrt{{n}+\mathrm{1}},{b}=\sqrt{{n}} \\ $$$${a}−{b}=\frac{\mathrm{1}}{\:\sqrt{{n}+\mathrm{1}}+\sqrt{{n}}} \\ $$$${and}\:{using}\:{sa}\:{m}\:,{sin}\left({a}\right)<\mathrm{1} \\ $$$$\Leftrightarrow\mid{cos}\left(\sqrt{{n}+\mathrm{1}}\right)−{cos}\left(\sqrt{{n}}\right)\mid\leqslant\mathrm{2}{sin}\left(\frac{\mathrm{1}}{\:\sqrt{{n}+\mathrm{1}}+\sqrt{{n}}}\right) \\ $$$$ \\ $$