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$f:\mathbb{R}\rightarrow\mathbb{R}$ 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 $f(a)=b$?

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    $\begingroup$ You write $f(a_n)\rightarrow b$. To me, that notation already implies that $\lim_{n\rightarrow\infty}f(a_n)$ exists (and equals $b$). $\endgroup$
    – Thorgott
    Dec 23, 2019 at 14:04
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    $\begingroup$ @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. $\endgroup$ Dec 23, 2019 at 14:05
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    $\begingroup$ 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 $f(x)=\begin{cases}x&\text{if }x\neq 3\\ 0&\text{if }x=3\end{cases}$ $\endgroup$
    – JMoravitz
    Dec 23, 2019 at 14:05
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    $\begingroup$ However, if the hypothesis were that for all $a_n\to a$ one has $f(a_n)\to b$, then yes, it implies that $f(a)=b$. $\endgroup$ Dec 23, 2019 at 14:06
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    $\begingroup$ Note that the hypotheses do not include that $f$ is a continuous function. $\endgroup$
    – JMoravitz
    Dec 23, 2019 at 14:06

2 Answers 2

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No. Consider $$ f(x)=\begin{cases} 0 & x<0 \\1 & else\end{cases} $$

and $b=0$ and $a_n=-\frac{1}{n}$.

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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 $a$.

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