Is my proof for $f(0)=1$ for a specific continuous function correct? Alright, I think I have found a much simpler proof to a question than the one I was provided with, and wanted to hear how it is inevitably incorrect.  
Let $f$ be a continuous function that you can always get the derivative of and that is always positive. Additionally,
$f'(0)=\lim\limits_{x\rightarrow 0}\frac{f(x)-1}{x}$
prove $f(0)=1$  
I proved it as such:  
According to the definition of a continuous function:  
$\lim\limits_{x\rightarrow c}f(x)=f(c)$  
Therefore:  
$f'(0)=\lim\limits_{x\rightarrow c}\frac{f(x)-1}{x}=\frac{f(c)-1}{c}\Rightarrow c\cdot f'(0)=f(c)-1$  
Let $c=0:$  
$f(0)-1=0\Rightarrow f(0)=1 \blacksquare$

The solution is not correct. By definition,
$$f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x}=\lim_{x\to 0}\frac{f(x)-1}{x}$$
so,
$$\lim_{x\to 0}\frac{f(x)-f(0)}{x}-\lim_{x\to 0}\frac{f(x)-1}{x}=0\to \lim_{x\to 0}\frac{f(0)-1}{x}=0\quad (1)$$
but 
$$\lim_{x\to 0}\frac{c}{x}=c\cdot \lim_{x\to 0}\frac{1}{x}$$
Doesn't exist if $c\ne 0$ and it is a constant number. 
Then if $(1)$ is true you must have $f(0)=1$.
P.S.: Your solution in not correct because when you write $c\cdot f'(0)=f(c)-1$ you are assuming that $c\ne 0$. 
