# A limit with limit zero everywhere must be zero somewhere [duplicate]

I wish to know if the following is true:

Let $$f : [\alpha, \beta]\to \mathbb R$$ be a function so that $$\lim_{x\to x_0} f(x) = 0$$ for all $$x_0 \in [\alpha, \beta]$$. Then $$f(x) = 0$$ for some $$x\in [\alpha, \beta]$$.

The Thomae's function $$f: [0,1]\to \mathbb R$$

$$f(x) =\begin{cases} 1/q & \text{if }x= p/q\in \mathbb Q, \\ 0 & \text{otherwise.}\end{cases}$$

leads me to the above question. The Thomae function has limit zero everywhere, althought it is nonzero in $$\mathbb Q$$. I think I can take any countable dense subset $$D\subset [\alpha, \beta]$$ and construct a function which is nonzero in $$D$$ but limit equals zero everywhere. But I can't think of a function that is nonzero everywhere but has zero limit everywhere.

The number of points $$x\in[\alpha,\beta]$$ at which $$|f(x)|>1/n$$, for a fixed $$n\in\mathbb{N}$$, has to be finite. Otherwise, since the interval is compact, they accumulate somewhere and at that point the limit wouldn't be zero.
Therefore, the points $$x\in[\alpha,\beta]$$ at which $$|f(x)|\neq0$$ is countable, but $$[\alpha,\beta]$$ isn't (unless $$\alpha=\beta$$ but that case follows directly).