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 $\rightarrow 0$

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. 
