Question: Let $f: \mathbb{R} \to \mathbb{R} $ be an increasing function. Suppose there are sequences $(x_n)$ and $(y_n)$ such that $x_n<0<y_n$ for all $n\geq 1$ and $f(y_n)-f(x_n) \to 0$ as $n \to \infty$. Prove that $f$ is continuous at $0$.
Attempt: We consider two arbitrary sequences $(a_n)$ with $a_n>0$ which converges to $0$ and $(b_n)$ with $b_n <0$ which converges to $0$ also.
We choose monotone subsequences $(a_{r_n})$ and $(b_{s_n})$. They both converge to $0$ [$(a_{r_n})$ must be monotone decreasing and the other one M.I.] . Again, being a monotone function, $f(0-)$ and $f(0+)$ both exist. Further, we have $f(b_{s_1})\leq f(b_{s_2})\leq...\leq f(b_{s_n})\leq f(a_{r_n})\leq f(a_{r_{n-1}})\leq ...\leq f(a_{r_1})$.
Now, we consider the Nested Intervals: $I_t=[f(b_{s_t}),f(a_{r_t})]$.
By the given condition $|I_t| \to 0$, as $t\to \infty $.
$\therefore f(0-),f(0+)\in \displaystyle \bigcap_{t=1}^{\infty}{I_t}$
Thereby, $f(0-)=f(0+)$. Again, the function being M.I, $f(0-)\leq f(0)\leq f(0+)$. Hence, $f(0-)=f(0)=f(0+)$
Did I miss anything? Kindly justify the proof. It will be of great help.
Thank You.