TITLE: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function. Which of the following conditions ensures that $\lim _{x \rightarrow 2} f(x)=\pi$?
QUESTION [0 upvotes]: Let  $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function. Which of the following conditions ensures that  $\lim _{x \rightarrow 2} f(x)=\pi$?
A) For each $\epsilon>0$ exists $\bar{n} \in \mathbb{N}$, such that $|f(2-1 / n)-\pi|<\epsilon$ for each $n \geq \bar{n}$
B)$ \lim _{n \rightarrow \infty} f(2-1 / n)=\pi$
My intuition tells me that a) and b) is right.  But the solution says that both are wrong. I just don't understand why. Is there perhaps a counterexample ?

REPLY [3 votes]: Let $f(2-1/n)=\pi$ for all $n$ and $f(x)=0$ for all $x$ not of the form $2-1/n$, this is a counterexample because you can take a sequence $(q_{n})$ of irrational points such that $q_{n}\rightarrow 2$, but $f(q_{n})=0$ for all $n$.
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