Limit definition of a derivative proof? Suppose that $f$ is a function with the properties: $f$ is differentiable everywhere, $f(x+y)=f(x)f(y)$, $f(0)≠0$, $f'(0)=1$. I need to learn how to use limit definition of the derivative to show $f'(x) = f(x)$ for all values of $x$.
I have: 
$$\lim \limits_{h \to 0} \frac{f(x+h)-f(x)}{h}$$
$$\lim \limits_{y \to 0} \frac{f(x+y)-f(x)}{h}$$ 
$$\lim \limits_{y \to 0} \frac{f(x)f(y)-f(x)}{h}$$ 
$$\lim \limits_{y \to 0} \frac{f(x)(1)-f(x)}{h}$$ 
$$\lim \limits_{y \to 0} \frac{f(x)-f(x)}{h}$$ 
...indeterminate form $\frac{0}{0}$
 A: First proof that $f(0)=1$
$$f(0+0)=f(0)f(0)$$
$$f(0)=f(0)f(0) \rightarrow f(0)=1$$
Indeed
$$f(nx)=f(\underbrace{x+\dots +x}_{n \small \mbox{ times}})=\underbrace{f(x)\times \dots \times f(x)}_{n \small \mbox{ times}}=[f(x)]^n $$
$$f(0x)=f(0)=[f(x)]^0=1$$
So you have 
$$f'(x)=\lim \limits_{h \to 0} \frac{f(x)f(h)-f(x)}{h}=\bigg (\lim \limits_{h \to 0} \frac{f(h)-1}{h} \bigg )f(x)$$
you can find out that 
$$f'(0)=\bigg (\lim \limits_{h \to 0} \frac{f(h)-1}{h}\bigg )f(0)=\lim \limits_{h \to 0} \frac{f(h)-1}{h}=1$$
So 
$$f'(x)=f'(0)f(x)=f(x)$$
A: You are making an error at the very first step, which unfortunately invalidates everything that comes afterwards, even though some of it is logically correct.  You can't substitute $y$ for $h$ everywhere except the denominator.  Actually, there is no need to introduce $y$ at all; you can just work with $h$:$$f'(x)=\lim_{h\to0}{f(x+h)-f(x)\over h}=\lim_{h\to0}{f(x)f(h)-f(x)\over h}=f(x)\lim_{h\to0}{f(h)-1\over h}$$
Can you continue from here?
A: From the definition of derivative, $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.$$  Since you are also told that $f$ satisfies the property $$f(x+y) = f(x)f(y)$$ for all $x, y$, it follows that $$f(x+h) = f(x)f(h),$$ hence $$f'(x) = \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h} = \lim_{h \to 0} f(x) \frac{f(h) - 1}{h} = f(x) \lim_{h \to 0} \frac{f(h) - 1}{h}.$$  The first step is simply applying the rule you were given.  The second factors our $f(x)$ from the numerator.  The third step uses the fact that $f(x)$ is not a function of $h$, thus it can be factored out of the limit.
The next step is to suppose that the quantity $$\lim_{h \to 0} \frac{f(h) - 1}{h}$$ exists.  If so, then clearly it does not depend on the choice of $x$, since we factored out $f(x)$.  So if it exists, let's call it some constant $c$.  What you must show is that if $f$ satisfies the other properties you were given, then we must have $c = 1$.  This part I have left as an exercise for the reader.
A: Hint: $f(0)=f(0+0)=f(0).f(0) \rightarrow f(0)=1$ because $f(0)\neq 0$
