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Question Number 174876 by infinityaction last updated on 13/Aug/22
let f:[0,1]→R be a continuous function ditermine (with appropriate justification) the following limit: lim_(n→∞) ∫_0 ^1 nx^n f(x)dx
$$\:\:\mathrm{let}\:{f}:\left[\mathrm{0},\mathrm{1}\right]\rightarrow\mathbb{R}\:{be}\:{a}\:\mathrm{continuous} \\ $$$$\:\mathrm{function}\:\mathrm{ditermine}\:\left(\mathrm{with}\:\mathrm{appropriate}\right. \\ $$$$\left.\mathrm{justification}\right)\:\mathrm{the}\:\mathrm{following}\: \\ $$$$\:\:\mathrm{limit}:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\int_{\mathrm{0}} ^{\mathrm{1}} {nx}^{{n}} {f}\left({x}\right){dx} \\ $$
Answered by TheHoneyCat last updated on 13/Aug/22
Sorry if it sounds rude, but your question is either meaningless or trivial when you write: lim_(x→∞) ∫_0 ^1 nx^n f(x)dx if you have a signle x it does′t mean anything now if you mean: (x≠x) lim_(x→∞) ∫_0 ^1 nx^n f(x)dx the limit is useless so the result is ∫_0 ^1 nx^n f(x)dx if you mean: lim_(x→∞) ∫_0 ^1 nx^n f(x)dx or lim_(x→∞) ∫_0 ^1 nx^n f(x)dx it′s also very trivial I think you probably meant lim_(n→∞) ∫_0 ^1 nx^n f(x)dx f beeing Continuous on an interval, it is bounded let F ∈R_+ ∀ε<1 ∫_0 ^ε nx^n Fdx=(n/(n+1))ε^(n+1) F (1) and ∫_0 ^1 nx^n Fdx=(n/(n+1))F (2) So we can use (2) to get that the integral cannot be bigger that the maximal value of F nor smaler than the smallest. because (2)→_(n→∞) F while (1) shows that the values smaler that the part of f on values smaler than 1 has no importance. because (1)→_(n→∞) 0 By considering increasingly bigger ε we find lim_(n→∞) ∫_0 ^1 nx^n f(x)dx=f(1) _□
$$\mathrm{Sorry}\:\mathrm{if}\:\mathrm{it}\:\mathrm{sounds}\:\mathrm{rude},\:\mathrm{but}\:\mathrm{your}\:\mathrm{question}\:\mathrm{is} \\ $$$$\mathrm{either}\:\mathrm{meaningless}\:\mathrm{or}\:\mathrm{trivial} \\ $$$$\mathrm{when}\:\mathrm{you}\:\mathrm{write}: \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\int_{\mathrm{0}} ^{\mathrm{1}} {nx}^{{n}} {f}\left({x}\right){dx} \\ $$$$\mathrm{if}\:\mathrm{you}\:\mathrm{have}\:\mathrm{a}\:\mathrm{signle}\:{x}\:\mathrm{it}\:\mathrm{does}'\mathrm{t}\:\mathrm{mean}\:\mathrm{anything} \\ $$$$ \\ $$$$\mathrm{now}\:\mathrm{if}\:\mathrm{you}\:\mathrm{mean}:\:\left({x}\neq{x}\right) \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\int_{\mathrm{0}} ^{\mathrm{1}} {nx}^{{n}} {f}\left({x}\right){dx} \\ $$$$\mathrm{the}\:\mathrm{limit}\:\mathrm{is}\:\mathrm{useless}\:\mathrm{so}\:\mathrm{the}\:\mathrm{result}\:\mathrm{is} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {nx}^{{n}} {f}\left({x}\right){dx} \\ $$$$ \\ $$$$\mathrm{if}\:\mathrm{you}\:\mathrm{mean}: \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\int_{\mathrm{0}} ^{\mathrm{1}} {nx}^{{n}} {f}\left({x}\right){dx} \\ $$$$\mathrm{or}\: \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\int_{\mathrm{0}} ^{\mathrm{1}} {nx}^{{n}} {f}\left({x}\right){dx} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{also}\:\mathrm{very}\:\mathrm{trivial} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{you}\:\mathrm{probably}\:\mathrm{meant} \\ $$$$\mathrm{lim}_{{n}\rightarrow\infty} \int_{\mathrm{0}} ^{\mathrm{1}} {nx}^{{n}} {f}\left({x}\right)\mathrm{d}{x}\: \\ $$$$ \\ $$$$ \\ $$$${f}\:\mathrm{beeing}\:\mathrm{Continuous}\:\mathrm{on}\:\mathrm{an}\:\mathrm{interval},\:\mathrm{it}\:\mathrm{is} \\ $$$$\mathrm{bounded} \\ $$$$\mathrm{let}\:{F}\:\in\mathbb{R}_{+} \:\forall\epsilon<\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\epsilon} {nx}^{{n}} {F}\mathrm{d}{x}=\frac{{n}}{{n}+\mathrm{1}}\epsilon^{{n}+\mathrm{1}} {F}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{and}\:\int_{\mathrm{0}} ^{\mathrm{1}} {nx}^{{n}} {F}\mathrm{d}{x}=\frac{{n}}{{n}+\mathrm{1}}{F}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\mathrm{So}\:\mathrm{we}\:\mathrm{can}\:\mathrm{use}\:\left(\mathrm{2}\right)\:\mathrm{to}\:\mathrm{get}\:\mathrm{that}\:\mathrm{the}\:\mathrm{integral}\: \\ $$$$\mathrm{cannot}\:\mathrm{be}\:\mathrm{bigger}\:\mathrm{that}\:\mathrm{the}\:\mathrm{maximal}\:\mathrm{value}\:\mathrm{of}\:\mathrm{F} \\ $$$$\mathrm{nor}\:\mathrm{smaler}\:\mathrm{than}\:\mathrm{the}\:\mathrm{smallest}. \\ $$$${because}\:\left(\mathrm{2}\right)\underset{{n}\rightarrow\infty} {\rightarrow}{F} \\ $$$$\mathrm{while}\:\left(\mathrm{1}\right)\:\mathrm{shows}\:\mathrm{that}\:\mathrm{the}\:\mathrm{values}\:\mathrm{smaler}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{part}\:\mathrm{of}\:{f}\:\mathrm{on}\:\mathrm{values}\:\mathrm{smaler}\:\mathrm{than}\:\mathrm{1}\:\mathrm{has}\:\mathrm{no}\:\mathrm{importance}. \\ $$$${because}\:\left(\mathrm{1}\right)\underset{{n}\rightarrow\infty} {\rightarrow}\mathrm{0} \\ $$$$\mathrm{By}\:\mathrm{considering}\:\mathrm{increasingly}\:\mathrm{bigger}\:\epsilon\:\mathrm{we}\:\mathrm{find} \\ $$$$\mathrm{lim}_{{n}\rightarrow\infty} \int_{\mathrm{0}} ^{\mathrm{1}} {nx}^{{n}} {f}\left({x}\right)\mathrm{d}{x}={f}\left(\mathrm{1}\right)\:\:\:\:_{\Box} \\ $$
Commented by infinityaction last updated on 13/Aug/22
yes sir it is lim_(n→∞) thanks
$${yes}\:{sir}\:{it}\:{is}\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\: \\ $$$${thanks} \\ $$
Commented by TheHoneyCat last updated on 23/Aug/22
your welcome ;•)
$$\mathrm{your}\:\mathrm{welcome} \\ $$$$\left.;\bullet\right) \\ $$

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