TITLE: Proving a limit using another limit
QUESTION [6 upvotes]: Let $f(x)$ be a functions that's defined at some neighbourhood of $0$
$$\lim\limits_{x \to 0} \frac{f(x)}{x} = 3$$

Prove that:
  $$
\lim_{x \to 0}\frac{f(3x)}{\ln(1+4x)} = 2.25
$$

I really don't know what to in order to convert the given limit $$\lim\limits_{x \to 0} \frac{f(x)}{x} = 3$$
and use it in the expression. I know I can do:
$$
\lim_{x \to 0}\frac{f(3x)}{3x}\frac{3x}{\ln(1+4x)}
$$
But how will that help me?

REPLY [10 votes]: You may write, as $x$ is near $0$, $x\neq0$,
$$
\frac{f(3x)}{\ln(1+4x)} =\frac{f(3x)}{3x}\frac{3x}{\ln(1+4x)}=\frac{f(3x)}{3x}\frac{4x}{\ln(1+4x)} \frac{3}{4}
$$
then use
$$
\lim_{x \to 0} \frac{f(3x)}{3x} = 3
$$
and
$$
\lim_{x \to 0} \frac{\ln(1+4x)}{4x}= \lim_{u \to 0} \frac{\ln(1+u)}{u}=1
$$
to conclude that the desired limit is $\displaystyle \frac{3\times 3}{4} =\frac94=2.25.$

REPLY [6 votes]: You're on the correct way:
$$
\lim_{x \to 0}\frac{f(3x)}{3x}\frac{3x}{\ln(1+4x)}=\lim_{x \to 0}\frac{f(3x)}{3x}\lim_{x \to 0}\frac{3x}{\ln(1+4x)}=\lim_{y \to 0}\frac{f(y)}{y}\lim_{x \to 0}\frac{3x}{4x}=3\times \frac34=\frac94
$$