# Limit of inverse functions

This is a question of mine that arose recently.

Let's have a real-valued function $f$ with domain $\mathbb{R}$, that is strictly decreasing and bijective on $\mathbb{R}$. Also, we have that $\lim_\limits{x\to -\infty}{f(x)}=+\infty$ and $\lim_\limits{x\to +\infty}{f(x)}=-\infty$. Prove that $\lim_\limits{x\to +\infty}{f^{-1}(x)}=-\infty$.

So, my questions are:

1) How to prove this, without infimum and supremum and without using the $\epsilon-\delta$ definition of a limit at all?

2) Do we have to have both $\lim_\limits{x\to -\infty}{f(x)}=+\infty$ and $\lim_\limits{x\to +\infty}{f(x)}=-\infty$ to conclude the result requested to prove, or is only one of those limits required?

3) Can we deduce more things about other limits and special properties of $f^{-1}(x)$, e.g. about $\lim_\limits{x\to -\infty}{f^{-1}(x)}$?

(I think that the above conclusions all seem very interesting from a mathematical aspect!)

• You can only talk about $f^{-1}$ when $f$ is a bijection, with the current conditions, this is not true. Mar 29 '16 at 17:51
• Do you assume $f$ to be continuous? Mar 29 '16 at 17:52
• strictly monotonic functions are invertible? Mar 29 '16 at 17:53
• @Jason If $f$ is not continuous, then $f^{-1}$ does not exists. Mar 29 '16 at 17:54
• @Hetebrij: That's not entirely true. Rather, if $f$ is strictly monotonic but not continuous, then $f$ is invertible, but the domain of its inverse is not all of $\Bbb R.$ Mar 29 '16 at 17:56

From the comments, I use the following "definition" for $$\lim_{x\to \infty} f(x) = -\infty$$: $$\forall K \in \mathbb R : \exists x\in\mathbb R : f(x) < K.$$ In general it is not the same, but for decreasing functions it is. $$\lim_{x\to -\infty} f(x) = \infty$$ is "defined" analogously.

There is nothing fancy here.

1. Notice that $$f^{-1}$$ is strictly decreasing as for $$y_1,y_2\in f(\mathbb R)$$ with $$y_1 = f(x_1) > f(x_2) = y_2$$ we have $$x_1 < x_2$$, otherwise it would contradict that $$f$$ is strictly decreasing.

2. Let $$K\in\mathbb R$$. Then, from $$\lim_{x\to -\infty} f(x) = \infty$$ there exists some $$x\in\mathbb R$$ with $$f(x) > f(K)$$, thus $$f^{-1}(y) < f^{-1}(f(K)) = K$$ for $$y=f(x)$$. That is $$\lim_{y\to \infty} f^{-1}(y) = -\infty$$.

On the other hand from $$\lim_{x\to \infty} f(x) = -\infty$$ there exists some $$\tilde x\in\mathbb R$$ with $$f(\tilde x) < f(K)$$, thus $$f^{-1}(\tilde y) > f^{-1}(f(K)) = K$$ for $$\tilde y = f(\tilde x)$$. That is, $$\lim_{y\to-\infty} f^{-1}(y) = \infty$$.

• May you, if you have the kindness, provide a similar proof using the $\epsilon-\delta$ austere limit definition? Just underneath the current proof. Also, thank you very much!!
– user171110
Mar 29 '16 at 21:42
• @Jason: It is basically the same. The $\epsilon$-$\delta$ definition for $\lim_{x\to\infty} f(x) = -\infty$ would be $\forall K\in\mathbb R : \exists x_0\in \mathbb R : \forall x > x_0 : f(x) < K$. Mar 29 '16 at 21:44