Limit of inverse function

Question

I need some help with the following question. Let $f$ be a real-valued one-to-one function with domain $(a-1, a+1)$. Let $\lim_{x\to a} f(x) = L$. Prove or disprove: $\lim_{y \to L} f^{-1}(y)=a$.

Answer 1

Let $I = (-1..1)$. We’ll construct a one-to-one function $f \colon I → I$ such that $\lim_{x → 0} f(x) = 0$ and $\lim_{x → -1} f(x) = 0 = \lim_{x → 1} f(x)$.

This is how $f$ looks like: $$f \colon I → I, \quad f(x) = \begin{cases} f_l(x) \quad &\text{for $x ∈ (-1..-1/2)$} \\ f_c(x) \quad &\text{for $x ∈ [-1/2..1/2]$} \\ f_r(x) \quad &\text{for $x ∈ (1/2..1)$}\\ \end{cases} $$

Now we need to define $f_l$, $f_c$ and $f_r$ appropriately.

For $x ∈ [-1..1]$, let’s look at balls $B_n(x) = \{y ∈ I; |x-y| < 1/n$} and their successive differences $C_n(x) = B_{n}(x) \setminus B_{n+1}(x)$. These are rings! So we’re splitting balls around $x$ into rings which get tinier and tinier the closer they become to $x$.

We will now create bijections: \begin{align*} l_n &\colon C_{n+1}(-1) → C_{3n - 1}(0)\\ c_n &\colon C_{n}(0) → C_{3n}(0), \\ r_n &\colon C_{n+1}(1) → C_{3n + 1}(0) \end{align*} (With $l_n$ and $r_n$ the codomain is $C_{n+1}(\pm 1)$ instead of $C_n(\pm 1)$ because we only need to cover $B_2(\pm 1)$ in the end.) These will yield exactly the one-to-one functions we need \begin{align*} f_l &= \bigsqcup_{n ∈ ℕ} l_n \colon B_2(-1) → I\\ f_c &= \bigsqcup_{n ∈ ℕ} c_n \colon B_1(0) → I \\ f_r &= \bigsqcup_{n ∈ ℕ} r_n \colon B_2(1) → I \\ \end{align*} This works since $B_k (x) = \bigsqcup_{n = k}^{∞} C_n(x)$. Well, actually, there are yet few technicalities: One should set $C_∞(x) = \{x\}$ for $x ∈ I$ as well as $f_c (0) = 0$ and $f_c (1/2) = 1/2$ and $f_c (-1/2) = 1/2$.

So what are $l_n$, $c_n$ and $r_n$? Just let them be (affine) linear.

Now $f$ is one-to-one and obviously $\lim_{x → 0} f(x) = 0$ and $\lim_{x → -1} f(x) = 0 = \lim_{x → 1} f(x)$. But you can’t say anything about $\lim_{y → 0} f^{-1}(y)$, can you?