$\lim_{x \to0} \frac{f(bx)}{bx} = \lim_{x \to 0} \frac{f(x)}{x}$ proof verification

Question

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 $$\lim_{x \to 0} \frac{f(bx)}{bx} =l$$

Answer 1

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 $\lim_{x \to 0} \frac{f(bx)}{bx} =l$.

Answer 2

Setting $bx:=y$ we see that, when $x\to 0$, $y\to 0$ too. Then $$\lim_{x\to 0}\frac {f(bx)}{bx}=\lim_{y\to 0}\frac {f(y)}{y}.$$ By alternating the symbols $y,x$ you get the desired form.