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I want to show that if $f:(a,b)\to \mathbb R$ is a strictly monotone function and if $x_0\in (a,b)$ such that there exist two sequences $(a_n)$ and $(b_n)$ with $a_n<x_0<b_n$ and $\lim\limits_{n\to \infty}(f(b_n)-f(a_n))=0$, then $f$ is continuous at $x_0$.

By monotonicity we can say that $\lim\limits_{n\to \infty}f(a_n)=f(x_0)=\lim\limits_{n\to \infty}f(b_n)$. Let $(x_n)$ be a sequence in $(a,b)$ such that $x_n\to x$. Here I got stuck. How to show that $f(x_n)\to f(x_0)$? Please help!

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We may assume that $f$ is strictly increasing. (Otherwise, replace $f$ with $-f$.)

Let $\epsilon > 0$. Since $\lim_{n \to \infty} (f(b_n) - f(a_n)) = 0$, there is some $N$ such that $|f(b_N) - f(a_N)| < \epsilon$.

As $a_N < x_0 < b_N$ and $x_n \to x_0$, there is some $M$ such that $a_N < x_n < b_N$ for all $n \geq M$.

Since $f$ is strictly increasing, we have the inequalities $$f(a_N) < f(x_0) < f(b_N) \qquad (1)$$ and, for $n \geq M$, $$f(a_N) < f(x_n) < f(b_N) \qquad (2)$$ Since (1) is equivalent to $$-f(b_N) < -f(x_0) < -f(a_N) \qquad (3)$$ we can add (2) and (3) to obtain $$-(f(b_N) - f(a_N)) < f(x_n) - f(x_0) < f(b_N) - f(a_N)$$ which is the same as $$|f(x_n) - f(x_0)| < f(b_N) - f(a_N) < \epsilon$$ which holds for all $n \geq M$. As $\epsilon > 0$ was arbitrary, we conclude that $f(x_n) \to f(x_0)$ as desired.

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    $\begingroup$ By the way, we don't really need $f$ to be strictly monotone. Nonstrict monotonicity is sufficient. In that case, all of the $<$ starting at the line labeled (1) become $\leq$, but that won't affect the outcome. $\endgroup$
    – user169852
    Aug 25, 2016 at 17:34

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