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 $\rightarrow 0$

If 
$$
\lim \frac{f(x)}{g(x)}=0
$$
and
$$
\lim g(x)=0,
$$
then
$$
\lim f(x) = \lim \frac{f(x)}{g(x)} \cdot g(x) =0.
$$
