- #1

Felafel

- 171

- 0

## Homework Statement

Let ##f:\mathbb{R}\to \mathbb{R}## a monotone function sucht that

## \displaystyle \lim_{x \to +\infty} \frac{f(2x)}{f(x)}=1##

show that for all c>0, we have

##\displaystyle \lim_{x \to +\infty} \frac{f(cx)}{f(x)}=1##

I think I'm almost there. Does it look okay to you? also, is it valid for 0<c<1 or just for c>1?

thank you very much

## The Attempt at a Solution

For the definition of limit to infinity:

##\forall \epsilon >0## ##\exists S>0## ##:##

##|f(x)-l|<\epsilon## ##\forall x>S##

##\displaystyle \lim_{x \to +\infty} \frac{f(2x)}{f(x)}=1## ##\Rightarrow## ##|\frac{f(2x)}{f(x)}-1|<\epsilon$ $\forall x>S##

which means

##f(x)(-\epsilon+1)<f(2x)<(\epsilon+1)f(x)## (I see it's monotonically decreasing, and so 1 is the infimum)

But if ##\forall \epsilon>0## i get ##-\epsilon f(x)+f(x)<f(2x)<\epsilon f(x)+f(x)##

Being ##\epsilon \to 0## ##\Rightarrow## ##|f(2x)-f(x)|=0##

And so:

##f(2x) \leq (1+\epsilon)f(x)##

##f(3x) \leq (1+\epsilon)f(2x)##

##f(3x) \leq (1+\epsilon)^2f(x)##

##1 \leq f(cx) \leq (1+\epsilon)^{c-1} f(x)##

##\epsilon \to 0## ##\Rightarrow## ##1 \leq f(cx) \leq f(x)## and, for the squeeze rule:

##\displaystyle \lim_{x \to +\infty} f(cx)=1## so

##\displaystyle \lim_{x \to +\infty} \frac{f(cx)}{f(x)}=1##