Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 116166 by Ar Brandon last updated on 01/Oct/20

∀n∈N^∗ , suppose u_n =(5sin(1/n^2 )+(1/5)cos n)^n Prove that lim_(n→+∞) u_n =0

$$\forall{n}\in\mathbb{N}^{\ast} ,\:\mathrm{suppose}\:{u}_{{n}} =\left(\mathrm{5sin}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}}\mathrm{cos}\:{n}\right)^{{n}} \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}}{u}_{{n}} =\mathrm{0} \\ $$

Answered by mindispower last updated on 01/Oct/20

since sin((1/n^2 ))→0⇒∃N such ∀n≥N 0≤sin((1/n^2 ))<(1/(10)) ∀x∈R −1≤cos(x)≤1 ⇒∀n≥N 0−(1/5)≤5sin((1/n^2 ))+((cos(n))/5)≤(5/(10))+(1/5)=(7/(10)) ∀n≥N −(1/5)≤5sin((1/n^2 ))+((cos(n))/5)≤(7/(10)) ⇒−((1/5))^n ≤(5sin((1/n^2 ))+((cos(n))/5))^n ≤((7/(10)))^n since −((1/5))^n →0,((7/(10)))^n →0 we get our Ansewr

$${since}\:{sin}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)\rightarrow\mathrm{0}\Rightarrow\exists{N}\:{such}\:\forall{n}\geqslant{N} \\ $$$$\mathrm{0}\leqslant{sin}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)<\frac{\mathrm{1}}{\mathrm{10}} \\ $$$$\forall{x}\in\mathbb{R}\:\:\:\:−\mathrm{1}\leqslant{cos}\left({x}\right)\leqslant\mathrm{1} \\ $$$$\Rightarrow\forall{n}\geqslant{N} \\ $$$$\mathrm{0}−\frac{\mathrm{1}}{\mathrm{5}}\leqslant\mathrm{5}{sin}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)+\frac{{cos}\left({n}\right)}{\mathrm{5}}\leqslant\frac{\mathrm{5}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{5}}=\frac{\mathrm{7}}{\mathrm{10}} \\ $$$$\forall{n}\geqslant{N} \\ $$$$\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{5}}\leqslant\mathrm{5}{sin}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)+\frac{{cos}\left({n}\right)}{\mathrm{5}}\leqslant\frac{\mathrm{7}}{\mathrm{10}} \\ $$$$\Rightarrow−\left(\frac{\mathrm{1}}{\mathrm{5}}\right)^{{n}} \leqslant\left(\mathrm{5}{sin}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)+\frac{{cos}\left({n}\right)}{\mathrm{5}}\right)^{{n}} \leqslant\left(\frac{\mathrm{7}}{\mathrm{10}}\right)^{{n}} \\ $$$${since}\:−\left(\frac{\mathrm{1}}{\mathrm{5}}\right)^{{n}} \rightarrow\mathrm{0},\left(\frac{\mathrm{7}}{\mathrm{10}}\right)^{{n}} \rightarrow\mathrm{0} \\ $$$${we}\:{get}\:{our}\:{Ansewr} \\ $$$$ \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com