Why-lim-n-x-n-u-n-0-When-u-n-is-bounded-

Question Number 204805 by York12 last updated on 27/Feb/24

$$\mathrm{Why}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left({x}^{{n}} {u}_{{n}} \right)=\mathrm{0}\:,\mathrm{When}\:{u}_{{n}} \:\mathrm{is}\:\mathrm{bounded}\: \\ $$

Commented by mr W last updated on 27/Feb/24

only true if ∣x∣<1. proof: when ∣x∣<1, lim_(x→∞) x^n =0 say a≤u_n ≤b ax^n ≤x^n u_n ≤bx^n 0=lim_(n→∞) ax^n ≤lim_(n→∞) x^n u_n ≤lim_(n→∞) bx^n =0 ⇒lim_(n→∞) x^n u_n =0

$${only}\:{true}\:{if}\:\mid{x}\mid<\mathrm{1}. \\ $$$${proof}: \\ $$$${when}\:\mid{x}\mid<\mathrm{1},\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}^{{n}} =\mathrm{0} \\ $$$${say}\:{a}\leqslant{u}_{{n}} \leqslant{b} \\ $$$${ax}^{{n}} \leqslant{x}^{{n}} {u}_{{n}} \leqslant{bx}^{{n}} \\ $$$$\mathrm{0}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{ax}^{{n}} \leqslant\underset{{n}\rightarrow\infty} {\mathrm{lim}}{x}^{{n}} {u}_{{n}} \leqslant\underset{{n}\rightarrow\infty} {\mathrm{lim}}{bx}^{{n}} =\mathrm{0} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}{x}^{{n}} {u}_{{n}} =\mathrm{0} \\ $$

Commented by justenspi last updated on 27/Feb/24

Thank you sir may I ask for analysis or calculud book recommendation

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{may}\:\mathrm{I}\:{ask}\:{for}\:\mathrm{analysis}\:\mathrm{or}\:\mathrm{calculud} \\ $$$$\mathrm{book}\:\mathrm{recommendation} \\ $$

Commented by justenspi last updated on 27/Feb/24