Let $f(x)$ be a function $(0,\infty) \to R$ and for every $x,y$ in the domain we have: $$f(xy) = f(x) + f(y)$$

It is like logarithm but we don't know the exact form of the function. we know it is differentiable at x=1. Now we want to show it is differentiable at its domain and its derivative is $f'(x) = \frac{1}{x} f'(1)$.


I can find that $f(1) = 0$ and $f(x/y) = f(x) - f(y)$ So we have:

$f'(1) = \lim_{h\to 0} \frac{f(1+h) - f(1)}{h} = \lim_{h\to 0}\frac{f(1+h)}{h} $


$f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h\to 0} \frac{f(\frac{x+h}{x})}{h} = \lim_{h\to 0} \frac{f(1 + h/x)}{h}$.

I know I can solve it with a simple change of variables $h \to 0 ~~~\rightarrow~~~~ h/x \to 0$, But I want to know is there any way to solve this without changing variable? Maybe using definition of limit?

  • 1
    $\begingroup$ @kimchilover it is stated explicitly that $f$ is differentiable at $x=1$. $\endgroup$ – Umberto P. Nov 26 '18 at 18:44
  • $\begingroup$ @kimchilover it is said in the body of the question that it is differentiable at 1. It is not my deduction. $\endgroup$ – amir na Nov 26 '18 at 18:49
  • $\begingroup$ OK, I understand now. $\endgroup$ – kimchi lover Nov 26 '18 at 19:05

You pointed out that $f'(1) = \displaystyle \lim_{h \to 0} \frac{f(1+h)}{h}$ exists using the fact that $f$ is differentiable at $x=1$.

A simple substitution shows that $\displaystyle \lim_{h \to 0} \frac{f(1+h/x)}{h/x} = f'(1)$ for all $x > 0$.

The above statement follows from a basic $\epsilon$ argument. Let $\epsilon > 0$ be given. There exists $\delta > 0$ with the property that $$0 < |h| < \delta \implies \left| \frac{f(1+h)}{h} - f'(1) \right| < \epsilon.$$ Thus $$0 < |h| < x \delta \implies 0 < \left| \frac hx \right| < \delta \implies \left| \frac{f(1+h/x)}{h/x} - f'(1) \right| < \epsilon$$ so that $$\frac{f(1+h/x)}{h/x} \to f'(1).$$

Now let $x > 0$ be arbitrary. Then $$f(x+h) - f(x) = f(x(1 + h/x)) - f(x) = f(1 + h/x)$$ so that $$\frac{f(x+h) - f(x)}{h} = \frac{f(1 + h/x)}{h} = \frac 1x \frac{f(1 + h/x)}{h/x}$$ and consequently $$\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \frac 1x \lim_{h \to 0} \frac{f(1 + h/x)}{h/x} = \frac 1x \cdot f'(1).$$

  • $\begingroup$ As I said in the question, I know it can be solved like this, But in this solution it is assumed that $h\to0$ means $h/x \to 0$. but someone told me when you don't know that the function is continuous, you can't simply substitute variables (i.e. h changed to h/x) . I want to know can it be proved just by definition of limit ($\epsilon , \delta$)? $\endgroup$ – amir na Nov 26 '18 at 18:47
  • $\begingroup$ Please see the edited answer. $\endgroup$ – Umberto P. Nov 26 '18 at 18:49
  • $\begingroup$ @amirna When $h\to 0$ and $x\ne 0$ then $\frac{h}{x}\to 0$. It is an easy exercise on $\epsilon-\delta$: $\left|\frac{h}{x}\right|<\epsilon$ iff $|h|<\epsilon|x|=\delta$. $\endgroup$ – A.Γ. Nov 26 '18 at 18:51
  • $\begingroup$ I saw your answer and I know it is completely valid. However, I want to know whether this question can be solved with definition of limit or not. I mean getting $\epsilon,\delta$ from f'(1) (because we know it exists) and then changing them somehow for using in the $f'(x)$ definition/ $\endgroup$ – amir na Nov 26 '18 at 18:54
  • $\begingroup$ @A.Γ. I saw your comment now, it seems fine but I think maybe it must be $|h/x| < \delta '$. because actually $\epsilon$ is used in $|f(x) - L| < \epsilon$ part of the definition and not in the boundary of the variable. $\endgroup$ – amir na Nov 26 '18 at 18:57

Set $y=1$ to verify that $f(1)=0$. Now when $h\to 0$ \begin{align} \frac{f(x+h)-f(x)}{h}&=\frac{f((1+h/x)x)-f(x)}{h}=\frac{f(1+h/x)+f(x)-f(x)}{h}=\\ &=\frac{f(1+h/x)}{h/x}\cdot \frac{1}{x}=\frac{f(1+h/x)-f(1)}{h/x}\cdot\frac{1}{x}\to\ ? \end{align}


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