# Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function. Which of the following conditions ensures that $\lim _{x \rightarrow 2} f(x)=\pi$?

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 ?

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 $$n$$.