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Question Number 127452 by snipers237 last updated on 29/Dec/20

$$Leta\in {\mathbb{R}}^{\ast },\underset{n\to \mathrm{\infty }}{\lim }{(asin\left(\frac{1}{n^2}\right)+\frac{1}{a}cos\left(n\right))}^n=0$$

Answered by Ar Brandon last updated on 29/Dec/20

$${\mathrm{u}}_{\mathrm{n}}={(\mathrm{asin}\frac{1}{{\mathrm{n}}^2}+\frac{1}{\mathrm{a}}\mathrm{cosn})}^{\mathrm{n}}$$$$\mid \sin \left(\frac{1}{{\mathrm{n}}^2}\right)\mid ⩽1\wedge \mid \mathrm{cosn}\mid ⩽1$$$$\Rightarrow \mid \mathrm{asin}\frac{1}{{\mathrm{n}}^2}+\frac{1}{\mathrm{a}}\mathrm{cosn}\mid ⩽\mathrm{a}\mid \sin \frac{1}{{\mathrm{n}}^2}\mid +\frac{1}{\mathrm{a}}\mid \mathrm{cosn}\mid ⩽\mathrm{a}+\frac{1}{\mathrm{a}}$$$$\Rightarrow \mid {\mathrm{u}}_{\mathrm{n}}\mid ⩽{(\mathrm{a}+\frac{1}{\mathrm{a}})}^{\mathrm{n}}.\mathrm{But}\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{(\mathrm{a}+\frac{1}{\mathrm{a}})}^{\mathrm{n}}=+\mathrm{\infty }$$$$\mathrm{Since}\underset{\mathrm{n}\to \mathrm{\infty }}{\lim sin}\frac{1}{{\mathrm{n}}^2}=0,\mathrm{\exists }N\in {\mathbb{N}}^{\ast }\mathrm{\forall }\mathrm{n}\in {\mathbb{N}}^{\ast }\wedge \mathrm{n}⩾N,$$$$-\frac{1}{\mathrm{a}}⩽\mathrm{asin}\frac{1}{{\mathrm{n}}^2}⩽\frac{1}{\mathrm{a}}.\mathrm{Thus}\mathrm{\forall }\mathrm{n}⩾N,$$$$-\frac{2}{\mathrm{a}}⩽\mathrm{asin}\frac{1}{{\mathrm{n}}^2}+\frac{1}{\mathrm{a}}\mathrm{cosn}⩽\frac{2}{\mathrm{a}}$$$$\mathrm{where}\mid \mathrm{asin}\frac{1}{{\mathrm{n}}^2}+\frac{1}{\mathrm{a}}\mathrm{cosn}\mid ⩽\frac{2}{\mathrm{a}}.\mathrm{We}\mathrm{deduce}\mathrm{that}\mathrm{\forall }\mathrm{n}⩾N,$$$$\mid {\mathrm{u}}_{\mathrm{n}}\mid ⩽{\left(\frac{2}{\mathrm{a}}\right)}^{\mathrm{n}}\mathrm{and}\mathrm{since}\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\left(\frac{2}{\mathrm{a}}\right)}^{\mathrm{n}}=0,$$$$\text{Double subscripts: use braces to clarify}$$

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