Proving Continuity of a Function Using the Definition of Differentiability

[sinClair]

Homework Statement

Suppose f:(0,\infty)->R and f(x)-f(y)=f(x/y) for all x,y in (0,\infty) and f(1)=0. Prove f is continuous on (0,\infty) iff f is continuous at 1.

Homework Equations

I think I ought to use these defn's of continuity: f continuous at a iff f(x)->f(a) as x->a
or f is cont at a iff for Xn->a, f(Xn)-f(a) as Xn->\infty

The Attempt at a Solution

The forward direction is immediate. For the backwards direction, we want to show that f(x)->f(a) as x->a for a in (0,\infty). So since f cont at 1, f(x)->f(1)=0 as x->1. I tried to manipulate this but couldn't find a way to make x->a instead of x->1.

Then I used the other definition and let Xn=1+1/n and Yn=a(Xn)=a+(a/n). Now Yn->a so just want to show that f(Yn)->f(a) as n->\infty. But f(Yn)->a*0=0 as n->\infty...

I know I have to use f(x)-f(y)=f(x/y) somehow. So I went backwards: So I want to show that f(x)-f(a)->0 as x->a. So that means I want f(x/a)->0 as x->a. But now I don't see how to incorporate the fact that f is continuous at a.I know this is related to the log function but don't think this problem requires me to appeal that fact... Note Xn and Yn are sequences indexed by n (I'm noob at this latex).Thanks for helping.

 

[ircdan]

take any a in (0, inf), then a !=0 so x/a makes sense and we have f(x) - f(a) = f(x/a), so f(x) = f(x/a) + f(a), now take the limit as x->a. Note as x->a, x/a -> 1

 

[sinClair]

Thank you, Dan. It makes so much sense now.