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Let $f:\mathbb R\to \mathbb R$ be an invertible function such that $$\lim_{x\to a} f(x)=b$$

for some $a,b\in \mathbb R$.

Does it follow that $$\lim_{x\to b}f^{-1}(x)= a,$$

where $f^{-1}$ denotes the inverse function of $f$?

Edit: When I consider the $\epsilon,\delta$-definition of the limit, I feel that there should be an example that $\lim_{x\to b}f^{-1}(x)\neq a$ due to the fact that $\epsilon,\delta$-definition is not symmetric (for a given $\epsilon>0$, we find $\delta>0$ such that ....).

However, if we further assume that $f$ is cont., $$b=\lim_{x\to b}x=\lim_{x\to b}f\circ f^{-1}(x)=f(\lim_{x\to b} f^{-1}(x)).$$ It follows that $\lim_{x\to b} f^{-1}(x)=f^{-1}(b)=a$. Thus, one needs a discontinuous function to have a counter example. I wonder whether there is any simple function with this property.

@Floris Claassens'a answer shows that there are some "ugly functions" with this property.

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    $\begingroup$ Interesting question! Maybe you could use "Series Reversion" as suggested by the answer in this post: math.stackexchange.com/questions/2360037/… $\endgroup$
    – NoChance
    Oct 17, 2019 at 9:37
  • $\begingroup$ Also consider the generalized series for the inverse formula given by: en.wikipedia.org/wiki/… $\endgroup$
    – NoChance
    Oct 17, 2019 at 9:45
  • $\begingroup$ @TheSimpliFire: Could you explain why it is of topic? $\endgroup$
    – mesel
    Oct 20, 2019 at 9:54
  • $\begingroup$ @TheSimpliFire: I edited the question. $\endgroup$
    – mesel
    Oct 21, 2019 at 20:30
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    $\begingroup$ Great, have voted to reopen. $\endgroup$ Oct 22, 2019 at 6:39

2 Answers 2

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A counter-example with $a=b=0$ and $\lim_{x\to a}f(x)=f(a).$

Let $(b_n)_{n\in \Bbb N}$ be a strictly decreasing real sequence with $b_1=1/2$ and with $\lim_{n\to \infty}b_n=0.$

For $x\le 0$ let $f(x)=x.$

For $n\in \Bbb N$ let f map $(b_{n+1},b_n]$ bijectively onto $(b_{n+1},b_n).$ And let $f(n)=b_n.$

Let $f$ map $(b_1,\infty)\setminus \Bbb N$ bijectively onto $(b_1,\infty).$

Then $f:\Bbb R\to \Bbb R$ is a bijection, and $\lim_{x\to 0}f(x)=0=f(0).$

But $(b_n)_{n\in \Bbb N}$ converges to $0$ while $f^{-1} (b_n)=n, $ so $f^{-1}(x)$ does not converge as $x\to 0.$

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    $\begingroup$ A specific example of this: $$f(x) = \begin{cases} x & x, 1/x \notin \Bbb N_+\\ x/2 & 1/x \in \Bbb N_+\\ 1/x & x \in \Bbb (2N_+ -1)\\ x/2 & x\in 2\Bbb N_+\end{cases}$$ $\endgroup$ Oct 17, 2019 at 17:42
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    $\begingroup$ For totally incomprehensible reasons I have received 2 negative votes for this today. $\endgroup$ Oct 18, 2019 at 11:43
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    $\begingroup$ There is a group which attacks both question and it answers without any reason? $\endgroup$
    – mesel
    Oct 20, 2019 at 9:55
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    $\begingroup$ I hope the moderators of the sites get to figure out how to control the process of down voting. The simplest approach is to force a comment of 80 characters to be typed and a down-vote reason from a drop-down list, like not clear answer, wrong answer, etc. $\endgroup$
    – NoChance
    Oct 20, 2019 at 14:29
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    $\begingroup$ @NoChance . I rarely down-vote and when I do I say why. I think down-voting should require a reason and be subject to review, like edits. This could be done without compromising anonymity. $\endgroup$ Oct 22, 2019 at 1:32
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If $x=a$, consider the sequence $(a_{n})=(a+\frac{b-a}{2}\frac{1}{n})_{n\geq 1}$ and the sequence $(b_{n})(b+\frac{b-a}{2}\frac{1}{n})_{n\geq1}$. We define $f$ as follows $$f(x)=\begin{cases}b+\frac{b-a}{2}\frac{1}{2n-1}&\text{ if }x=a+\frac{b-a}{2}\frac{1}{n},\ n>0.\\ b+\frac{b-a}{2}\frac{1}{2n}&\text{ if }x=b+\frac{b-a}{2}\frac{1}{2n},\ n>0.\\2b-a+\frac{b-a}{2}\frac{1}{n}&\text{ if }x=b+\frac{b-a}{2}\frac{1}{2n-1},\ n>0.\\x+b-a&\text{ else}.\end{cases}$$ Note that $f$ is well-defined as for all $m,n\in\mathbb{N}$ we have \begin{align*}|a+\frac{b-a}{2}\frac{1}{m}-b-\frac{b-a}{2}\frac{1}{n}|&=|a-b+\frac{b-a}{2}\frac{n-m}{mn}|\geq|a-b|-|\frac{a-b}{2}||\frac{1}{m}-\frac{1}{n}|\\&\geq|a-b|-\frac{1}{2}|a-b|>0\end{align*} Using similar arguments we find that $$f^{-1}(x)=\begin{cases}a+\frac{b-a}{2}\frac{1}{n}&\text{ if }x=b+\frac{b-a}{2}\frac{1}{2n-1},\ n>0.\\b+\frac{b-a}{2}\frac{1}{2n} &\text{ if }x=b+\frac{b-a}{2}\frac{1}{2n},\ n>0.\\b+\frac{b-a}{2}\frac{1}{2n-1}&\text{ if }x=2b-a+\frac{b-a}{2}\frac{1}{n},\ n>0.\\x+a-b&\text{ else}.\end{cases}$$ is well-defined, and evidently $f^{-1}$ is the inverse of $f$. So $f$ is bijective.

Now let $\varepsilon>0$ and take $\delta=\min(\varepsilon,\frac{b-a}{2})$. For all $x\in\mathbb{R}$ with $|x-a|<\delta$ we have $|f(x)-f(a)|\leq|x+b-a-b|=|x-a|<\varepsilon$. So $\lim_{x\rightarrow a}f(x)=b$.

Furthermore note that $\lim_{n\rightarrow\infty}f^{-1}(b+\frac{b-a}{2}\frac{1}{2n})=b\neq a$, so $\lim_{x\rightarrow b}f^{-1}(x)\neq a$.

For $a=b$ one can actually (contrary to my previous claim) define a similar function. We define $$f(x)=\begin{cases}a+\frac{1}{2n}&\text{ if }x=a+\frac{1}{n},\ n>0.\\a+\frac{1}{2n-1}&\text{ if }x=a+2+\frac{1}{2n-1},\ n>0.\\a+2+\frac{1}{n}&\text{ if }x=a+2+\frac{1}{2n},\ n>0.\\x&\text{ else}.\end{cases}$$ Using similar arguments we find that $f$ is bijective $\lim_{x\rightarrow a}f(x)=a$, but $\lim_{n\rightarrow \infty}f^{-1}(a+\frac{1}{2n-1})=a+2\neq a$.

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  • $\begingroup$ your function is not well defined. $a+\dfrac{1}{n}$ can be equal to $b+\dfrac{1}{2n}$. Even if you corrected this part, you need to explain clearly your solution. $\endgroup$
    – mesel
    Oct 17, 2019 at 9:48
  • $\begingroup$ @mesel but for $a\neq b$ you can always take $n > N$ with $N$ large enough to make sure that $a+1/n\neq b+1/(2n)$. $\endgroup$
    – Surb
    Oct 17, 2019 at 9:50
  • $\begingroup$ @FlorisClaassens " if $𝑎=𝑏$ there is no function satisfying the conditions in the question ", how do you know that ? $\endgroup$
    – Surb
    Oct 17, 2019 at 9:50
  • $\begingroup$ @Surb It is only written that $n>0$ and $n>1$. As I wrote, even if this part is corrected, the solution should be explained. $\endgroup$
    – mesel
    Oct 17, 2019 at 9:52
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    $\begingroup$ Just made a quick fix to solve the issue surrounding a+\frac{1}{n}=b+\frac{1}{2n}$, I'll work on a slightly more thorough explanation, though I had hoped that the idea behind the function was clear. $\endgroup$ Oct 17, 2019 at 9:59

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