1. Jul 24, 2006

### JoAuSc

I was looking at the definition for a limit when I was wondering what would happen if you changed this

(f(x+h)-f(x))/h

to this

( f(x+h)/f(x) )^(1/h)

with h going to zero in both cases. I did a few calculations with the second limit on my graphing calculator and got this:

function limit
f(x) = x e^(1/x)
f(x) = x^3 e^(3/x)
f(x) = sin(x) e^(1/tax(x))

Does this limit have any kind of importance beyond just being an interesting limit? I was trying to come up with a limit which would measure the ratio a function increases by over a small distance rather than the difference (as for the derivative).

2. Jul 24, 2006

### StatusX

Try taking the log of it.

3. Jul 25, 2006

### JoAuSc

Thanks for the help.

After I posted last night I continued studying the limit and eventually realized that limit when applied to a function was equal to e^(d/dx ln(f(x))), which I thought was interesting because rather than Df we have L^-1 D L f, where D is d/dx, L is ln(x), and f is the function. It reminds me of something I read about normal subgroups, though I don't think that'd apply here.

4. Jul 26, 2006

### StatusX

Well, you are effectively conjugating differentiation by the log operation. One application would be that applying this operator repeatedly, the inner logs and exps cancel and you are left with something like (LDL-1)n = L Dn L-1. I don't know if this is useful, but maybe by picking the right operator/function L, you can do repeated differentiation of some functions more easily this way.