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Question Number 174876 by infinityaction last updated on 13/Aug/22

$$\mathrm{let}f:[0,1]\to \mathbb{R}bea\mathrm{continuous}$$$$\mathrm{function}\mathrm{ditermine}(\mathrm{with}\mathrm{appropriate}$$$$\mathrm{justification})\mathrm{the}\mathrm{following}$$$$\mathrm{limit}:\underset{n\to \mathrm{\infty }}{\lim }{\int }_0^1{nx}^{n}f\left(x\right)dx$$

Answered by TheHoneyCat last updated on 13/Aug/22

$$\mathrm{Sorry}\mathrm{if}\mathrm{it}\mathrm{sounds}\mathrm{rude},\mathrm{but}\mathrm{your}\mathrm{question}\mathrm{is}$$$$\mathrm{either}\mathrm{meaningless}\mathrm{or}\mathrm{trivial}$$$$\mathrm{when}\mathrm{you}\mathrm{write}:$$$$\underset{x\to \mathrm{\infty }}{\lim }{\int }_0^1{nx}^{n}f\left(x\right)dx$$$$\mathrm{if}\mathrm{you}\mathrm{have}\mathrm{a}\mathrm{signle}x\mathrm{it}{\mathrm{does}}^{\prime }\mathrm{t}\mathrm{mean}\mathrm{anything}$$$$$$$$\mathrm{now}\mathrm{if}\mathrm{you}\mathrm{mean}:(x\ne x)$$$$\underset{x\to \mathrm{\infty }}{\lim }{\int }_0^1{nx}^{n}f\left(x\right)dx$$$$\mathrm{the}\mathrm{limit}\mathrm{is}\mathrm{useless}\mathrm{so}\mathrm{the}\mathrm{result}\mathrm{is}$$$${\int }_0^1{nx}^{n}f\left(x\right)dx$$$$$$$$\mathrm{if}\mathrm{you}\mathrm{mean}:$$$$\underset{x\to \mathrm{\infty }}{\lim }{\int }_0^1{nx}^{n}f\left(x\right)dx$$$$\mathrm{or}$$$$\underset{x\to \mathrm{\infty }}{\lim }{\int }_0^1{nx}^{n}f\left(x\right)dx$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{also}\mathrm{very}\mathrm{trivial}$$$$\mathrm{I}\mathrm{think}\mathrm{you}\mathrm{probably}\mathrm{meant}$$$${\lim }_{n\to \mathrm{\infty }}{\int }_0^1{nx}^{n}f\left(x\right)\mathrm{d}x$$$$$$$$$$$$f\mathrm{beeing}\mathrm{Continuous}\mathrm{on}\mathrm{an}\mathrm{interval},\mathrm{it}\mathrm{is}$$$$\mathrm{bounded}$$$$\mathrm{let}F\in {\mathbb{R}}_{+}\mathrm{\forall }ϵ<1$$$${\int }_0^{ϵ}{nx}^{n}F\mathrm{d}x=\frac{n}{n+1}{ϵ}^{n+1}F\left(1\right)$$$$\mathrm{and}{\int }_0^1{nx}^{n}F\mathrm{d}x=\frac{n}{n+1}F\left(2\right)$$$$\mathrm{So}\mathrm{we}\mathrm{can}\mathrm{use}\left(2\right)\mathrm{to}\mathrm{get}\mathrm{that}\mathrm{the}\mathrm{integral}$$$$\mathrm{cannot}\mathrm{be}\mathrm{bigger}\mathrm{that}\mathrm{the}\mathrm{maximal}\mathrm{value}\mathrm{of}\mathrm{F}$$$$\mathrm{nor}\mathrm{smaler}\mathrm{than}\mathrm{the}\mathrm{smallest}.$$$$because\left(2\right)\underset{n\to \mathrm{\infty }}{\to }F$$$$\mathrm{while}\left(1\right)\mathrm{shows}\mathrm{that}\mathrm{the}\mathrm{values}\mathrm{smaler}\mathrm{that}$$$$\mathrm{the}\mathrm{part}\mathrm{of}f\mathrm{on}\mathrm{values}\mathrm{smaler}\mathrm{than}1\mathrm{has}\mathrm{no}\mathrm{importance}.$$$$because\left(1\right)\underset{n\to \mathrm{\infty }}{\to }0$$$$\mathrm{By}\mathrm{considering}\mathrm{increasingly}\mathrm{bigger}ϵ\mathrm{we}\mathrm{find}$$$${\lim }_{n\to \mathrm{\infty }}{\int }_0^1{nx}^{n}f\left(x\right)\mathrm{d}x=f\left(1\right){}_{◻}$$

Commented by infinityaction last updated on 13/Aug/22

$$yessiritis\underset{n\to \mathrm{\infty }}{\lim }$$$$thanks$$

Commented by TheHoneyCat last updated on 23/Aug/22

$$\mathrm{your}\mathrm{welcome}$$$$;\bullet )$$

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