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 $\rightarrow 0$

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.
