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 $c=0:$  
$f(0)-1=0\Rightarrow f(0)=1 \blacksquare$

Your proof is incorrect, because you can't write
$$
\lim_{x\to0}\frac{f(x)-1}{x}=\frac{f(c)-1}{c}
$$
Intuitively, $f(0)=1$ because otherwise the limit
$$
\lim_{x\to0}\frac{f(x)-1}{x}
$$
cannot exist finite.
More rigorously,
$$
f(0)=\lim_{x\to0}f(x)=
\lim_{x\to0}\left(1+\frac{f(x)-1}{x}x\right)=
1+\lim_{x\to0}\frac{f(x)-1}{x}\cdot\lim_{x\to0}x
$$
because the limits in the last term exist finite.
