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 $\rightarrow 0$

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 $f(1/3) \
eq 0$ 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 $f(x_{0}) = 0$.