Existence of solutions of the equation with a limit. Let f be continuous function on [0,1] and
$$\lim_{x→0} \frac{f(x + \frac13) + f(x + \frac23)}{x}=1$$
Prove that exist $x_{0}\in[0,1]$ which satisfies equation $f(x_{0})=0$
I suppouse that the numerator should approach $0$ which would implicate that for 
x near $0$ $f(x + \frac13)$ would be of opposite sign then $f(x + \frac23)$ or both be $0$. 
Then by intermediate value theorem we would know that there is $x_{0}\in[\frac13,\frac23]$ which fulfill  $f(x_{0})=0$
Yet, I have no idea how to prove that numerator $\rightarrow 0$
 A: Clearly we can see that $$\lim_{x \to 0}f(x + 1/3) + f(x + 2/3) = \lim_{x \to 0}x\cdot\frac{f(x + 1/3) + f(x + 2/3)}{x} = 0 \cdot 1 = 0$$ and by continuity of $f$ we see that this implies $f(1/3) + f(2/3) = 0$. If $f(1/3) = 0$ then we are done. If $f(1/3) \neq 0$ then both $f(1/3)$ and $f(2/3)$ are of opposite signs and hence by intermediate value theorem there is an $x_{0} \in (1/3, 2/3)$ for which $f(x_{0}) = 0$.
A: Hint: If the numerator doesn't go to $0$, can the limit goes to 1? At first suppose that there exists $\lim_{x\to 0} (f(x+1/3)+f(x+2/3))$ and isn't zero.
A: First, prove that the numerator should be zero. You can do this by contradiction. Prove that if $$\lim_{x\to 0} f(x+\frac13) + f(x+\frac23)\neq 0$$
then the original limit cannot exist.
Now, you can use continuity to show that $f(\frac23) + f(\frac13)=0$, and then use a well known theorem to finish your proof.
A: If 
$$
\lim \frac{f(x)}{g(x)}=0
$$
and
$$
\lim g(x)=0,
$$
then
$$
\lim f(x) = \lim \frac{f(x)}{g(x)} \cdot g(x) =0.
$$
A: I have a slightly different idea:
The codintion indicates that: $f(x+1/3) + f(x+2/3)$ is an equivalent infinitesimal to the function $x$.., which means that $f(x+1/3)+f(x+2/3)$ is equivalent to $ o(x)$, when $x$ is infinitely small...
So $f(x+1/3)+f(x+2/3)\approx a_0 + a_1x$ and $a_0 = 0$, $a_1 = 0$ or $1$.
$a_1=0$ apparently leads to the result..
if $a_1 = 1$, it $f(x+1/3)+f(1+2/3)=x$. so $f(1/3)+f(2/3)=0$, which also leads to the conclusion. 
