# Limits of the inverse function

Consider a real function $$f$$ which is defined, continuous and strictly monotonic on a given interval $$I$$, yielding an inverse $$f^{-1}:f(I) \rightarrow I$$.

Can one assert that for any $$a$$ and $$b$$ in the closures of $$I$$ and $$f(I)$$ respectively, the following implication holds?

$$\lim\limits_{x \to a}f(x) = b \Longrightarrow \lim\limits_{y \to b}f^{-1}(y) = a$$

($$a$$ and $$b$$ might be infinite)

Thanks for any replies.

• Yes what you think is true and can be easily proved by applying the definition of limit and continuity. You have to understand that $f^{-1}$ is also continuous and strictly monotone on interval $f(I)$. – Paramanand Singh Feb 22 at 4:47

Start considering the case $$a\in I$$ and $$b\in f(I)$$, so that they are both finite. By continuity of $$f$$ we have $$b=\lim_{x\rightarrow a} f(x)=f(a)$$ and thus by continuity of $$f^{-1}$$ we have $$\lim_{y\rightarrow b} f^{-1}(y)= f^{-1}(b)=f^{-1}(f(a))=a.$$ Remark. Note that in case $$a$$ is an endpoint of $$I$$ (and thus $$f(a)$$ and endpoint of $$f(I)$$) some of the previous limits are one-sided limits.
The other cases can be proved by direct verification of the definition of limit. For instance, suppose $$f$$ to be strictly increasing on $$I=[a,+\infty)$$ and such that $$\lim_{x\rightarrow +\infty} f(x)=b$$. Note that $$f^{-1}$$ is also strictly increasing. We want to prove that $$\lim_{y\rightarrow b^-} f^{-1}(y) = +\infty$$. Let $$M>0$$ and let $$\delta:=b-f(M)$$. Then $$y>b-\delta=f(M) \iff f^{-1}(y)>M,$$ and this is the definition of $$\lim_{y\rightarrow b^-} f^{-1}(y) = +\infty$$.