Prove that $\lim_{x \to 0} \frac {f(3x)}{\ln(1+4x)} = \frac94$ if $\lim_{x \to 0} \frac {f(x)}{x} = 3$ I need to prove the following:
$f(x)$ defined on a neighborhood of $x=0$ such that:
$$ \lim_{x \to 0} \frac  {f(x)}{x} = 3 $$
I need to prove that:
$$ \lim_{x \to 0} \frac {f(3x)}{\ln(1+4x)} = 2.25  $$
Basically, what I did is:
$$ \lim_{x \to 0} \frac {f(3x)}{\ln(1+4x)} = \lim_{x \to 0} \frac {f(3x)}{x} \cdot \frac{x}{\ln(1+4x)} = \lim_{x \to 0} \frac {f(3x)}{x} \cdot \frac{4x}{\ln(1+4x)} \cdot \frac{1}{4}   $$
Now I'm stuck here with 2 questions:
1) Can I say that since $x \to 0$ it is hold that:
$$ \lim_{x \to 0} \frac{f(x)}{x} = \lim_{x \to 0} \frac{f(3x)}{x} $$ 
If so, then I can simply switch this expression to $3$.
2) The other question is about the $\ln$ expression, I know the identity that:
$$ \lim_{x \to 0} \frac {\ln(1+x)}{x} = 1 $$
But in my expression I need to switch the denominator with the numerator in order to use this Identity. How can I do so?
I'm not sure about any of my steps so please correct me where I'm wrong and don't just offer your own solution.
Thank you
 A: For the first question note that $y := 3x \to 0$ for $x \to 0$, but you have to replace $x$ by $3x$ everywhere, so 
$$ \frac{f(3x)}{x} = \frac{f(3x)}{3x} \cdot 3 = \frac{f(y)}{y} \cdot 3 \to 3 \cdot 3 = 9 $$
For the second problem, note that (with $z := 4x$ we have $z \to 0$ for $x \to 0$) it holds: 
$$ 
\frac{4x}{\log(1 + 4x)} = \frac 1{\log(1+z)/z} \to \frac 11 = 1 $$
So we have 
$$ \frac{f(3x)}{x} \cdot \frac{4x}{\log(1 + 4x)} \cdot \frac 14 \to 9 \cdot 1 \cdot \frac 14 = \frac 94 $$
A: Another observation is : 
$$\lim_{x \to 0 }\frac{f(x)}{x}=3 \to f(x) \sim 3x+o(x^2)\\
\lim_{x \to 0 }\frac{f(3x)}{\ln(1+4x)}=\lim_{x \to 0 }\frac{3(3x)+o(x^3)}{\ln(1+4x)}=\\\lim_{x \to 0 }\frac{9x+o(x^2)}{4x+o(x^2)}=\lim_{x \to 0 }\frac{9x}{4x}=\frac{9}{4}=2.25$$
in your way :take $3x=a \to  x=\frac{a}{3} \to 0$
$$\lim_{x \to 0 }\frac{f(3x)}{ln(1+4x)}=\\
\lim_{a \to 0 }\frac{f(a)}{ln(1+4\frac{a}{3})}=\\
\lim_{a \to 0 }\frac{f(a)}{a}\frac{a}{ln(1+4\frac{a}{3})}=\\
3.\lim_{a \to 0 }\frac{a}{ln(1+4\frac{a}{3})}=\\
3.\lim_{a \to 0 }\frac{a}{4\frac{a}{3}}=2.25$$
A: To prove this sort of limit, you will need to go back to the $\delta-\epsilon$ definition but first, it is helpful to derive, in a correct but non-rigorous manner, what the answer will be, because you can then use those steps to determine what $\delta$ must be in terms of $\epsilon$.
Your first step trick is good:
$$
\lim_{x\to 0} \frac{f(3x)}{\ln(1+4x)} = \lim_{x\to 0} \left(\frac{f(3x)}{3x}\right)
\left( \frac{3x}{\ln(1+4x)} \right) = \left(\lim_{x\to 0} \frac{f(3x)}{3x}\right)
\left( \lim_{x\to 0} \frac{3x}{\ln(1+4x)} \right)
$$
where the second step in that equation is only justified if each individual limit exists -- that is what I mean by a non-rigorous manner.
Then you can find those limits by changing variables to $y=3x$ and $u= \ln(1+4x)$  so that in the first limit, $x=y/3$ and in the second, $x=\frac{e^u-1}{4}$.
$$
\left(\lim_{x\to 0} \frac{f(3x)}{3x}\right)
\left( \lim_{x\to 0} \frac{3x}{\ln(1+4x)} \right)
=\left(\lim_{y\to 0} \frac{f(y)}{y}\right)
\left( \lim_{u\to 0} \frac34 \frac{e^u-1}{u} \right)= \frac94  \lim_{u\to 0} \frac{e^u-1}{u} 
$$
A: Hint: Why not write the limit as $$\lim_{x \to 0} \frac {f(3x)}{3x}\times \frac {4x}{\ln (1+4x)} \times \frac {3}{4}??$$
