Suppose that the function $f:\mathbb{R}\rightarrow\mathbb{R}$ is differentiable at $0$. For real number $a$, $b$, and $c$, with $c\neq 0$ prove that $$lim_{x\rightarrow 0}\frac{f(ax)-f(bx)}{cx}=[\frac{a-b}{c}]f'(0).$$
I have been toying around with some derivative algebra and the fact that if $f$ is differentiable then $f$ is continuous but I can't quite seem to work this one out.
$$\frac{f(ax)-f(bx)}{cx}=\frac{f(ax)-f(0)}{cx}-\frac{f(bx)-f(0)}{cx}=\frac ac\frac{f(ax)-f(0)}{ax}-\frac bc\frac{f(bx)-f(0)}{bx}$$ tends to
$$\frac acf'(0)-\frac bcf'(0).$$