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?

  • $\begingroup$ You could use a substitution to find the limit of the first factor and (shudder) L'Hôpital for the second. $\endgroup$ – David Mitra Dec 20 '14 at 11:21
  • $\begingroup$ I'm not allowed to use L'Hopital yet :\, but how will I substitute? it's $f(3x)$, not $f(x)$ $\endgroup$ – FigureItOut Dec 20 '14 at 11:22
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    $\begingroup$ $3x\rightarrow0$ if $x\rightarrow0$. You have $3x$ downstairs too. Setting $u=3x$, the limit is $\lim_{u\rightarrow0} {f(u)\over u}$. $\endgroup$ – 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 $\displaystyle \frac{3\times 3}{4} =\frac94=2.25.$


You're on the correct way:

$$ \lim_{x \to 0}\frac{f(3x)}{3x}\frac{3x}{\ln(1+4x)}=\lim_{x \to 0}\frac{f(3x)}{3x}\lim_{x \to 0}\frac{3x}{\ln(1+4x)}=\lim_{y \to 0}\frac{f(y)}{y}\lim_{x \to 0}\frac{3x}{4x}=3\times \frac34=\frac94 $$

  • $\begingroup$ How did you get rid of the $ln$ ? $\endgroup$ – FigureItOut Dec 20 '14 at 11:27
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    $\begingroup$ 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. $\endgroup$ – user63181 Dec 20 '14 at 11:29

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