If $\lim_{x \to a} f\left(x\right) = L$ then does $\lim_{x \to L} f^{-1}\left(x\right) = a$

Question

It seems intuitively true to be based upon logarithm and exponential, as well as tangent and inverse tangent, but is this always true?

Answer 1

Assuming that $f$ is only known to be bijective (so that $f^{-1}$ exists) the answer is no. Consider for example $f:[0,1]\cup(2,3]\to[0,2]$ with $f(x)=x$ if $x\in [0,1]$ and $f(x)=x-1$ if $x\in (2,3]$. Then it is clear that $f$ is bijective and satisfies $$\lim_{x\to 1}f(x)=1$$ However the inverse function $f^{-1}:[0,2]\to[0,1]\cup(2,3]$ is given by $f^{-1}(y)=y$ if $y\in [0,1]$ and $f(y)=y+1$ if $y\in (1,2]$. So the limit $$\lim_{y\to 1}f^{-1}(y)$$ doesn't exist because $\lim_{y\to1^+}f^{-1}(y)=2$ and $\lim_{y\to1^-}f^{-1}(y)=1$.
(Basically $f$ glues together the two intervals to one ($[0,2]$) while $f^{-1}$ tears that interval apart into two pieces and is therefore not continuous)
Edit: A positive result: If $f:I\to \Bbb R$ is defined on an interval $I$ (may be unbounded) and is continuous and injective with image $f(I)$, then the inverse function $f(I)\to I$ is also continuous.

Answer 2

I dont think this is necessarily true. In general if this were true we would call such a map homeomorphic. That is both function and its inverse is continuous. A sufficient condition for this to happen is the domain needs to be compact. You can construct non compact domains on which the inverse is not continuous.