Prove that if $\lim_{x \to 0} \frac{f(x)}{x} =l$ and $b \neq 0$,
$$\lim_{x \to0} \frac{f(bx)}{bx} = \lim_{x \to 0} \frac{f(x)}{x}$$
Given $$\lim_{x \to 0} \frac{f(x)}{x} =l, \; b\neq0$$
Suppose $$\lim_{x \to 0} \frac{f(bx)}{bx} =m$$
$$\Rightarrow \forall \epsilon > 0, \; \exists \delta>0 \;\text{such that } 0<|x|<\delta \Rightarrow \bigg|\frac{f(bx)}{bx}- m\bigg| < \epsilon$$
Let $u=bx$
$$\forall \epsilon > 0, \; \exists \delta>0 \;\text{such that } 0<\bigg|\frac{u}{b}\bigg|<\delta \Rightarrow \bigg|\frac{f(u)}{u}-m\bigg| < \epsilon$$
$$\forall \epsilon > 0, \; \exists \delta>0 \;\text{such that } 0<|u|<\delta|b| \Rightarrow \bigg|\frac{f(u)}{u}-m\bigg| < \epsilon$$
Letting $\delta'=\delta|b|$
$$\forall \epsilon > 0, \; \exists \delta'>0 \;\text{such that } 0<|u|<\delta' \Rightarrow \bigg|\frac{f(u)}{u}-m\bigg| < \epsilon$$
Which implies
$$\lim_{u \to 0}\frac{f(u)}{u}=m$$
which is eqivalent to
$$\lim_{x \to 0}\frac{f(x)}{x}=m$$
Since limits are unique and $\lim_{x \to 0} \frac{f(x)}{x} = l, \; l=m$
Therefore
$$\lim_{x \to 0} \frac{f(bx)}{bx} =l$$
Best Answer
Your proof seems fine, we can also proceed directly as follows without refer to unicity theorem, starting from the hypothesis $\lim_{x \to 0} \frac{f(x)}{x} =l$ since for $0<|x|<\delta$
$$ \bigg|\frac{f(x)}{x}- l\bigg| < \epsilon $$
then we have that for $0<|x|<\frac{\delta}{|b|} \iff 0<|bx|<\delta$
$$ \bigg|\frac{f(bx)}{bx}- l\bigg| < \epsilon $$
that is the thesis $\lim_{x \to 0} \frac{f(bx)}{bx} =l$.