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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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