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Let $f: \Bbb R \to \Bbb R$ and $x \in \Bbb R$. Suppose that $\lim_{y \to x+} f(y)$ exists as a real number. If there is an $r \in \Bbb R$ such that $$\lim_{y \to x+} f(y) > r$$ then there exists $n \in \Bbb N$ (dependind on $x$ and $r$) such that $$f(z)>r$$ whenever $x<z<x+1/n$.

I think this is intuitively obvious, but I'm trying to understand this in a formal way, maybe with the definition of limit of a function.

Jean-Sébastien
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  • This seems a bit awkward notationally, but otherwise fairly sound reasoning. Have you considered epsilon-delta calculus? – abiessu Sep 14 '13 at 23:16

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If $L$ is the limit, let $\epsilon = L - r$. Then by definition of a limit, there exists a $\delta > 0$ such that whenever $z \in (x, x + \delta$, we have $|f(z) - L| < \epsilon$. In particular, it follows that $f(z) > r$ for all such $z$.

Now just choose $n$ sufficiently large that $\frac{1}{n} < \delta$.