Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 146994 by KONE last updated on 17/Jul/21

$$u_n=cos\sqrt{n+1}-cos\sqrt{n}$$$$\underset{x\to +\mathrm{\infty }}{\lim }{u}_n=??$$

Answered by mathmax by abdo last updated on 17/Jul/21

$${\mathrm{u}}_{\mathrm{n}}=\cos (\sqrt{\mathrm{n}(1+\frac{1}{\mathrm{n}})}-\cos \left(\sqrt{\mathrm{n}}\right)$$$$=\cos \left(\sqrt{\mathrm{n}}\sqrt{1+\frac{1}{\mathrm{n}}}\right)-\cos \left(\sqrt{\mathrm{n}}\right)$$$$\sim \cos \left(\sqrt{\mathrm{n}}(1+\frac{1}{2\mathrm{n}})\right)-\cos \left(\sqrt{\mathrm{n}}\right)$$$$=\cos (\sqrt{\mathrm{n}}+\frac{1}{2\sqrt{\mathrm{n}}})-\cos \left(\sqrt{\mathrm{n}}\right)$$$$\mathrm{cosp}-\mathrm{cosq}=\mathrm{cosp}+\cos (\pi -\mathrm{q})=2\cos \left(\frac{\mathrm{p}+\pi -\mathrm{q}}{2}\right)\cos \left(\frac{\mathrm{p}-\pi +\mathrm{q}}{2}\right)$$$$=2\cos (\frac{\pi }{2}+\frac{\mathrm{p}-\mathrm{q}}{2})\cos (\frac{\mathrm{p}+\mathrm{q}}{2}-\frac{\pi }{2})=-2\sin \left(\frac{\mathrm{p}-\mathrm{q}}{2}\right)\cos \left(\frac{\mathrm{p}+\mathrm{q}}{2}\right)\Rightarrow$$$${\mathrm{u}}_{\mathrm{n}}\sim -2\sin \left(\frac{1}{4\sqrt{\mathrm{n}}}\right)\cos (\sqrt{\mathrm{n}}+\frac{1}{4\sqrt{\mathrm{n}}})\Rightarrow$$$$\mid {\mathrm{u}}_{\mathrm{n}}\mid ⩽2\mid \sin \left(\frac{1}{4\sqrt{\mathrm{n}}}\right)\mid \mathrm{due}\mathrm{to}\mid \cos (\sqrt{\mathrm{n}}+\frac{1}{4\sqrt{\mathrm{n}}})\mid ⩽1\mathrm{we}\mathrm{have}$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}\sin \left(\frac{1}{4\sqrt{\mathrm{n}}}\right)=0\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{u}}_{\mathrm{n}}=0$$

Answered by mathmax by abdo last updated on 17/Jul/21

$$\mathrm{autre}\mathrm{method}\mathrm{on}\mathrm{utilise}\mathrm{le}\mathrm{theoreme}\mathrm{des}\mathrm{accroissements}\mathrm{finis}\Rightarrow$$$$\Rightarrow \mathrm{\exists }\xi \in ]\sqrt{\mathrm{n}},\sqrt{\mathrm{n}+1}[\mathrm{tel}\mathrm{que}{\mathrm{u}}_{\mathrm{n}}=(\sqrt{\mathrm{n}+1}-\sqrt{\mathrm{n}})(-\frac{1}{2\sqrt{\mathrm{c}}}\sin \sqrt{\mathrm{c}})(\mathrm{f}\left(\mathrm{x}\right)=\cos \left(\sqrt{\mathrm{x}}\right))$$$$\Rightarrow {\mathrm{u}}_{\mathrm{n}}=-\frac{\sqrt{\mathrm{n}+1}-\sqrt{\mathrm{n}}}{2\sqrt{\mathrm{c}}}\sin \sqrt{\mathrm{c}}=-\frac{1}{2\sqrt{\mathrm{c}}(\sqrt{\mathrm{n}+1}+\sqrt{\mathrm{n}})}\sin \sqrt{\mathrm{c}}\Rightarrow$$$$\mid {\mathrm{u}}_{\mathrm{n}}\mid ⩽\frac{1}{2\sqrt{\mathrm{c}}(\sqrt{\mathrm{n}+1}+\sqrt{\mathrm{n}})}\to 0(\mathrm{n}\to +\mathrm{\infty })\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{u}}_{\mathrm{n}}=0$$

Answered by mindispower last updated on 17/Jul/21

$$f\left(x\right)=cos\left(\sqrt{x}\right),meanvalueth$$$$\Rightarrow \mathrm{\forall }(a,b){\mathrm{\exists }}_c\in ]a,b[such$$$$\Rightarrow f\left(a\right)-f\left(b\right)=f^{\prime }\left(c\right)(a-b)$$$$\Rightarrow cos\left(\sqrt{n+1}\right)-cos\left(\sqrt{n}\right)=-\frac{1}{2\sqrt{c}}sin\left(\sqrt{c}\right)(n+1-n)$$$$\Rightarrow \mid cos(\sqrt{n+1)}-cos\left(\sqrt{n}\right)\mid =\frac{1}{2\sqrt{c_n}}\mid sin\left(c_n\right)\mid$$$$usinc{c}_n\in ]n,n+1\lfloor ,\mid sin\mid <1$$$$\Rightarrow \mid cos\left(\sqrt{n+1}\right)-cos\left(\sqrt{n}\right)\mid ⩽\frac{1}{2\sqrt{n}}$$$$2ndway$$$$cos\left(a\right)-cos\left(b\right)=-2sin\left(\frac{a-b}{2}\right)sin\left(\frac{a+b}{2}\right)$$$$a=\sqrt{n+1},b=\sqrt{n}$$$$a-b=\frac{1}{\sqrt{n+1}+\sqrt{n}}$$$$andusingsam,sin\left(a\right)<1$$$$\iff \mid cos\left(\sqrt{n+1}\right)-cos\left(\sqrt{n}\right)\mid ⩽2sin\left(\frac{1}{\sqrt{n+1}+\sqrt{n}}\right)$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com