# $a_n\rightarrow a, f(a_n)\rightarrow b$ imply $f(a)=b$?

$$f:\mathbb{R}\rightarrow\mathbb{R}$$ is a function and $$a_n\rightarrow a, f(a_n)\rightarrow b$$. Then does the existence of $$\lim\limits_{n\rightarrow\infty}^{}f(a_n)$$ imply $$f(a)=b$$?

• You write $f(a_n)\rightarrow b$. To me, that notation already implies that $\lim_{n\rightarrow\infty}f(a_n)$ exists (and equals $b$). Dec 23, 2019 at 14:04
• @Thorgott The question is: Does the existence of the limit force said limit to be the value $f(a)$. The answer to that is negative. Dec 23, 2019 at 14:05
• Consider what it might mean if it didn't. Consider such an example where $f(a)\neq b$ despite $\lim\limits_{n\to\infty} f(a_n)=b$ and $\lim\limits_{n\to\infty}a_n=a$. Does this violate any of your hypotheses? Consider for explicit example something like $f(x)=\begin{cases}x&\text{if }x\neq 3\\ 0&\text{if }x=3\end{cases}$ Dec 23, 2019 at 14:05
• However, if the hypothesis were that for all $a_n\to a$ one has $f(a_n)\to b$, then yes, it implies that $f(a)=b$. Dec 23, 2019 at 14:06
• Note that the hypotheses do not include that $f$ is a continuous function. Dec 23, 2019 at 14:06

No. Consider $$f(x)=\begin{cases} 0 & x<0 \\1 & else\end{cases}$$
and $$b=0$$ and $$a_n=-\frac{1}{n}$$.
This is true if $$f$$ is continuous in $$a$$, since then, by definition if continuity, $$\forall\varepsilon>0\;\exists\delta > 0\;\forall y \in X: |a-y| < \delta \implies |f(a)-f(y)| < \varepsilon$$. Let $$a_n\rightarrow a$$ and choose an arbitrary $$\varepsilon > 0$$. Then, by definition of continuity, $$\exists\delta > 0\;\forall y \in X: |a-y| < \delta \implies |f(a)-f(y)| < \varepsilon$$. Since $$a_n$$ converges to $$a$$, it follows that $$\exists N \in \mathbb{N}\;\forall n \geq N: |a_n-a| < \delta$$ and so, $$|f(a_n)-f(a)| < \varepsilon$$. Ultimately, we have shown that $$\forall \varepsilon > 0\;\exists N = N(\delta) \in \mathbb{N}\;\forall n \geq N: |f(a_n)-f(a)| < \varepsilon$$, i.e. $$f(a_n)$$ converges to $$f(a)$$, which is what we wanted to prove. However, your statement is not true for every function $$f$$, in particular, it is true if and only if $$f$$ is continuous in $$a$$.