# Limit of an inverse function

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.

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

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 $$x\to 0.$$

• A specific example of this: $$f(x) = \begin{cases} x & x, 1/x \notin \Bbb N_+\\ x/2 & 1/x \in \Bbb N_+\\ 1/x & x \in \Bbb (2N_+ -1)\\ x/2 & x\in 2\Bbb N_+\end{cases}$$ Oct 17, 2019 at 17:42
• For totally incomprehensible reasons I have received 2 negative votes for this today. Oct 18, 2019 at 11:43
• There is a group which attacks both question and it answers without any reason? Oct 20, 2019 at 9:55
• 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. Oct 20, 2019 at 14:29
• @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. Oct 22, 2019 at 1:32

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 $$\lim_{n\rightarrow \infty}f^{-1}(a+\frac{1}{2n-1})=a+2\neq a$$.

• 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. Oct 17, 2019 at 9:48
• @mesel but for $a\neq b$ you can always take $n > N$ with $N$ large enough to make sure that $a+1/n\neq b+1/(2n)$.
– Surb
Oct 17, 2019 at 9:50
• @FlorisClaassens " if $𝑎=𝑏$ there is no function satisfying the conditions in the question ", how do you know that ?
– Surb
Oct 17, 2019 at 9:50
• @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. Oct 17, 2019 at 9:52
• 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. Oct 17, 2019 at 9:59