Does $\lim_{x\rightarrow0}\left(\frac{f(x) - f(0)}{x}\right) = \lim_{\frac{x}{3}\rightarrow0}\left(\frac{f(x) - f(0)}{x}\right)$? Given that $ f'(0) = 3$, I need to solve $$\lim_{x\rightarrow0}\left(\frac{f(3x) - f(0)}{x}\right)$$
Because I know that $\lim_{x\rightarrow0}\left(\frac{f(x) - f(0)}{x}\right) = 3$, I made the substitution $3x = h$. 
Thus, $$\lim_{x\rightarrow0}\left(\frac{f(3x) - f(0)}{x}\right) = 3 \lim_{\frac{h}{3}\rightarrow0}\left(\frac{f(h) - f(0)}{h}\right)$$
If, in fact the following is true, then I can conclude that $\lim_{x\rightarrow0}\left(\frac{f(3x) - f(0)}{x}\right) = 9$, and my work is done.
$$\lim_{x\rightarrow0}\left(\frac{f(x) - f(0)}{x}\right) = \lim_{\frac{x}{3}\rightarrow0}\left(\frac{f(x) - f(0)}{x}\right)\,?$$
 A: You are supposed to use here the substitution of limits:

Substitution in limits: Let $f$ be defined in a deleted neighborhood of $a$ and let $\lim_{x\to a} f(x) =L$. Further let $g$ be defined in a certain deleted neighborhood of $b$ such that $g(x) \neq a$ for all values of $x$ in this deleted neighborhood of $b$ and $\lim_{x\to b} g(x) =a$. Then $\lim_{x\to b} f(g(x)) =L$.

For the current question $a=b=0$ and $$F(x) =\frac{f(x) - f(0)}{x},G(x)=3x$$ Then we are given $\lim_{x\to 0}F(x)=3$ and hence $\lim_{x\to 0}F(G(x))=3$ ie $$\lim_{x\to 0}\frac{f(3x)-f(0)}{3x}=3$$ Multiplying the above by $3$ we can see that $$\lim_{x\to 0}\frac{f(3x)-f(0)}{x}=9$$ Also note that although the intended meaning of the notation $\lim_{x/3\to 0}$ is clear this is not a standard notation.
The rule of substitution is normally used without too much symbolism as follows
\begin{align*}
L&=\lim_{x\to 0}\frac{f(3x)-f(0)}{x}\\
&=\lim_{t\to 0}3\cdot\frac{f(t)-f(0)}{t}\text{ (putting } t=3x)\\
&=3\cdot 3=9
\end{align*} 
