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Question Number 204805 by York12 last updated on 27/Feb/24
Why lim_(n→∞) (x^n u_n )=0 ,When u_n  is bounded
$$\mathrm{Why}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left({x}^{{n}} {u}_{{n}} \right)=\mathrm{0}\:,\mathrm{When}\:{u}_{{n}} \:\mathrm{is}\:\mathrm{bounded}\: \\ $$
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
thank you
$${thank}\:{you} \\ $$
Commented by mr W last updated on 28/Feb/24
sorry, i know no book which i can  recommend.
$${sorry},\:{i}\:{know}\:{no}\:{book}\:{which}\:{i}\:{can} \\ $$$${recommend}. \\ $$
Commented by York12 last updated on 28/Feb/24
thank you.
$$\mathrm{thank}\:\mathrm{you}. \\ $$

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