# Limit of Maximum

• Nov 16th 2011, 01:40 AM
younhock
Limit of Maximum
Anyone can help me ?

QUESTION :
for all x in (0,1), n is Natural number

f_n (x) := max{1/x , n }

find lim f_n (x) when n -> infinity.
• Nov 16th 2011, 07:10 AM
HallsofIvy
Re: Limit of Maximum
Do you mean {1/n, n}? If so then you are looking at "max(1/2, 2)", "max(1/3, 3)", "max(1/4, 4)"...:"max{1/1000000, 1000000}", ... What are those numbers? What is the limit as n goes to infinity?

If you really mean {1/x, n} for fixed x it is essentially the same: "max(1/x, 2)", "max(1/x, 3)", "max(1/x 4)"...:"max{1/x 1000000}", where the first number is some fixed finite number.
• Nov 16th 2011, 07:31 AM
younhock
Re: Limit of Maximum
Quote:

Originally Posted by HallsofIvy
Do you mean {1/n, n}? If so then you are looking at "max(1/2, 2)", "max(1/3, 3)", "max(1/4, 4)"...:"max{1/1000000, 1000000}", ... What are those numbers? What is the limit as n goes to infinity?

If you really mean {1/x, n} for fixed x it is essentially the same: "max(1/x, 2)", "max(1/x, 3)", "max(1/x 4)"...:"max{1/x 1000000}", where the first number is some fixed finite number.

i had updated the question to make it clear.
I was interested to know what is the limit function which f_n approaching to .
• Nov 16th 2011, 12:37 PM
HallsofIvy
Re: Limit of Maximum
It's pretty straight forward if you write out a few terms in the sequence of functions.
For x in (0, 1) (in particular for x positive) 1/x> n when x< 1/n. So $f_n(x)= 1/x$ for x< 1/n and $f_n(x)= n$ for x> 1/n, In particular, $f_1(x)= 1/x$ for all x in (0, 1). $f_2(x)= 1/x$ for 0< x<= 1/2, $f_2(x)= 2$ for 1/2<= x< 1. $f_3(x)= 1/x$ for 0< x<= 1/3 and $f_3(x)= 3$ for 1/3<= x< 1. Do see what is happening? Given any x, there is eventually an integer N> 1/x so for n> N, $f_n(x)> N$.
• Nov 16th 2011, 06:30 PM
younhock
Re: Limit of Maximum
Quote:

Originally Posted by HallsofIvy
It's pretty straight forward if you write out a few terms in the sequence of functions.
For x in (0, 1) (in particular for x positive) 1/x> n when x< 1/n. So $f_n(x)= 1/x$ for x< 1/n and $f_n(x)= n$ for x> 1/n, In particular, $f_1(x)= 1/x$ for all x in (0, 1). $f_2(x)= 1/x$ for 0< x<= 1/2, $f_2(x)= 2$ for 1/2<= x< 1. $f_3(x)= 1/x$ for 0< x<= 1/3 and $f_3(x)= 3$ for 1/3<= x< 1. Do see what is happening? Given any x, there is eventually an integer N> 1/x so for n> N, $f_n(x)> N$.

i see, so we will get f_n(x) -> infinity eventually. Thanks yea.