THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

Consider $a,b\in \mathbb{R}$ such that $f(a)=f(b)$. The sequence $(x_n{)}_{n\ge 1}$ defined by $x_n=$ converges as $(f(x_n){)}_{n\ge 1}$ is convergent, concluding that $f$ is injective. $f$ is surjective by definition, $f$ is invertible.

Consider now $x_0\in \mathbb{R}$ and a sequence $(x_n{)}_{n\ge 1}$ with limit $x_0$. The sequence $(f(f^{-1}(x_n)){)}_{n\ge 1}$ is convergent, so the sequence $(f^{-1}(x_n){)}_{n\ge 1}$ is also. Remark that the sequence $(y_n{)}_{n\ge 1}$ given by $y_n=$ has also the limit $x_0$.

Thus the sequence $(f^{-1}(y_n){)}_{n\ge 1}$ is convergent, so $$\underset{n\to \infty }{\lim }{f}^{-1}(y_{2n})=\underset{n\to \infty }{\lim }{f}^{-1}(y_{2n-1}),$$ that is ${\lim }_{n\to \infty }{f}^{-1}(x_n)=f(x_0)$. As $x_0$ was arbitrary chosen $f^{-1}$ is continuous on $\mathbb{R}$ and $f$ is also.