Find $\lim_{x\to0} \frac{\log(1+3x)}{f(x)}$ given $\lim_{x\to0} \frac{f(x)}{\sin(x)} = 2$ It's assumed that $\lim_{x\to0} \frac{f(x)}{\sin(x)} = 2$. 

Find
  $$\lim_{x\to0} \frac{\log(1+3x)}{f(x)}$$ 

I don't think that it would work out by a random plugging. 
Let $f(x) = 2\sin(x).$
$$\lim_{x\to0}\frac{\log(1+3x)}{f(x)} = \lim_{x\to0}\frac{\log(1+3x)}{2\sin(x)} = [\log(1+3x) = 3x + O(x^2)] = \lim_{x\to0}\frac{3x+O(x^2)}{2\sin(x)} = \frac{3}{2}.$$
But what did I miss? I cannot find a way to prove whether these two limits are connected (or that the result is unique). 
 A: How about using these (well-known) limits:
$$\lim_{x\to 0} \frac{\ln (1+x)}{x}=1,\ \ \lim_{x\to 0} \frac{\sin x}{x}=1$$
and writing:
$$\lim_{x\to 0}\frac{\ln(1+3x)}{f(x)} = 3\lim _{x\to 0} \left[\frac{\ln (1+3x)}{3x}\cdot  \frac{x}{\sin x}\cdot \frac{\sin x}{f(x)}\right]$$
Can you end it from here?
A: Hint:
If the limits exist,
$$\lim_{x\to0} \frac{f(x)}{\sin(x)}\cdot\lim_{x\to0} \frac{\log(1+3x)}{f(x)}=\lim_{x\to0} \frac{f(x)}{\sin(x)}\frac{\log(1+3x)}{f(x)}=\lim_{x\to0} \frac{\log(1+3x)}{\sin(x)}.$$
A: With  equivalence of functions near $0$:
The hypothesis means that $f(x)\sim_0 2\sin x$
On the other hand $\sin x\sim_0 x$ and $\ln(1+x)\sim_0 x$, so
$$\frac{\log(1+3x)}{f(x)}\sim_0 \frac{3x}{2x}=\frac 32.$$
A: When $f(x) \ne 0; \sin x \ne 0$ we have 
$\frac {\log{1+3x}}{f(x)} = \frac {\log{1+3x}}{\sin x}\frac {\sin x}{f(x)}=\frac {\log{1+3x}}{\sin x}\cdot \frac 1{\frac {f(x)}{\sin x}}$
so if $\lim_{x\to 0}\frac {\log{1+3x}}{\sin x}$ exists than 
$\lim_{x\to 0}\frac {\log{1+3x}}{f(x)} = \lim_{x\to 0}\frac {\log{1+3x}}{\sin x}\cdot\frac 1{\lim_{x\to 0}\frac {f(x)}{\sin x}} = \frac 12\lim_{x\to 0}\frac {\log{1+3x}}{\sin x}$
And we can use L'hopital to solve $\lim_{x\to 0}\frac {\log{(1+3x)}}{\sin x}=\lim_{x\to 0} \frac {3\frac 1{1+3x}}{\cos x}= \frac {3\frac 11}{1}=3$
So $\lim_{x\to 0}\frac {\log{1+3x}}{f(x)}=\frac 12\lim_{x\to 0}\frac {\log{1+3x}}{\sin x}=\frac 32$.
====== old answer =====
You don't let $f(x) = 2\sin x$ but you note that as $\lim_{x\to 0} \frac {f(x)}{\sin x} = 2$ then for any $\lim_{x\to 0} h(x) = K$ then $\lim_{x\to 0} h(x)\frac {f(x)}{\sin x} = 2K$.
So if $\lim_{x\to 0} \frac {\log(1 +3x)}{f(x)}$ exists
then $\lim_{x\to 0} \frac {\log{(1 +3x)}}{f(x)}= \frac12\lim_{x\to 0} \frac {\log{(1 +3x)}}{f(x)}\frac {f(x)}{\sin x}=\frac 12\lim_{x\to 0}\frac {\log{(1+3x)}}{\sin x}$
And we can use L'hopital to solve $\frac 12\lim_{x\to 0}\frac {\log{(1+3x)}}{\sin x}=\frac 12\lim_{x\to 0} \frac {3\frac 1{1+3x}}{\cos x}= \frac 12\frac {3\frac 11}{1}=\frac 32$.
I suppose I skirted the issue as to whether it is possible for $\lim_{x\to 0} \frac {\log(1 +3x)}{f(x)}$ to NOT exist.  
But it must, as $\lim_{x\to 0} \frac {\log(1+3x)}{\sin x}=3$ and $\lim_{x\to 0}\frac {f(x)}{\sin x}=2$ do exist it must follow that $\frac 32 = \lim_{x\to 0}\frac {\frac {\log(1+3x)}{\sin x}}{ \frac{f(x)}{\sin x}}=\lim_{x\to 0}\frac {\log(1+3x)}{f(x)}$.
