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-

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

let f : [0, 1] \to R b e a continuous

function ditermine (with appropriate

justification) the following

limit : \lim_{n \to \infty} \int_{0}^{1} {n x}^{n} f (x) d x

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 \to \infty} \int_{0}^{1} {n x}^{n} f (x) d x

if you have a signle x it {does}^{'} t mean anything

now if you mean : (x \neq x)

\lim_{x \to \infty} \int_{0}^{1} {n x}^{n} f (x) d x

the limit is useless so the result is

\int_{0}^{1} {n x}^{n} f (x) d x

if you mean :

\lim_{x \to \infty} \int_{0}^{1} {n x}^{n} f (x) d x

or

\lim_{x \to \infty} \int_{0}^{1} {n x}^{n} f (x) d x

{it}^{'} s also very trivial

I think you probably meant

\lim_{n \to \infty} \int_{0}^{1} {n x}^{n} f (x) d x

f beeing Continuous on an interval, it is

bounded

let F \in R_{+} \forall ϵ < 1

\int_{0}^{ϵ} {n x}^{n} F d x = \frac{n}{n + 1} ϵ^{n + 1} F (1)

and \int_{0}^{1} {n x}^{n} F d x = \frac{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 .

b e c a u s e (2) \underset{n \to \infty}{\to} F

while (1) shows that the values smaler that

the part of f on values smaler than 1 has no importance .

b e c a u s e (1) \underset{n \to \infty}{\to} 0

By considering increasingly bigger ϵ we find

\lim_{n \to \infty} \int_{0}^{1} {n x}^{n} f (x) d x = f (1)_{}

Commented by infinityaction last updated on 13/Aug/22

y e s s i r i t i s \lim_{n \to \infty}

t h a n k s

Commented by TheHoneyCat last updated on 23/Aug/22

your welcome

; ∙)