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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.

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    $\begingroup$ 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) $. $\endgroup$ – Paramanand Singh Feb 22 at 4:47
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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 $\lim_{y\rightarrow b^-} f^{-1}(y) = +\infty$.

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