# Proving a limit using another limit

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?

• You could use a substitution to find the limit of the first factor and (shudder) L'Hôpital for the second. – David Mitra Dec 20 '14 at 11:21
• I'm not allowed to use L'Hopital yet :\, but how will I substitute? it's $f(3x)$, not $f(x)$ – FigureItOut Dec 20 '14 at 11:22
• $3x\rightarrow0$ if $x\rightarrow0$. You have $3x$ downstairs too. Setting $u=3x$, the limit is $\lim_{u\rightarrow0} {f(u)\over u}$. – David Mitra Dec 20 '14 at 11:25

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 $\displaystyle \frac{3\times 3}{4} =\frac94=2.25.$
$$\lim_{x \to 0}\frac{f(3x)}{3x}\frac{3x}{\ln(1+4x)}=\lim_{x \to 0}\frac{f(3x)}{3x}\lim_{x \to 0}\frac{3x}{\ln(1+4x)}=\lim_{y \to 0}\frac{f(y)}{y}\lim_{x \to 0}\frac{3x}{4x}=3\times \frac34=\frac94$$
• How did you get rid of the $ln$ ? – FigureItOut Dec 20 '14 at 11:27
• I used the Taylor expansion $$\ln(1+4x)\sim_0 4x$$ and if you don't allowed to use it you can use the L'hôpital's rule. – user63181 Dec 20 '14 at 11:29