Originally Posted by

**eXist** Well the problem reads: Suppose $\displaystyle f: (a, b) \rightarrow R$ and $\displaystyle \lim_{x\to\infty}f(x_n)$ exists for every sequence $\displaystyle x_n$ (where n goes from $\displaystyle 1 \to \infty$) in $\displaystyle (a, b)$ such that $\displaystyle \lim_{x\to\infty}x_n = a$. Prove that $\displaystyle \lim_{x\to a^+}f(x_n)$ exists.

I'm going to assume that $\displaystyle \lim_{x\to a^+}f(x_n)$ does not exist and prove it by contradiction. So then we still have $\displaystyle \lim_{x\to\infty}x_n = a$. But since the sequence $\displaystyle x_n$ lies in the interval $\displaystyle (a, b)$ the $\displaystyle \lim_{x\to\infty}x_n = a$ can't exist because we assumed that right hand limit: $\displaystyle \lim_{x\to a^+}f(x_n)$ does not exist, so the limit as anything approaches a in the interval $\displaystyle (a, b)$ can't exist. But that is a contradiction since we are given $\displaystyle \lim_{x\to\infty}x_n = a$.

I'm not sure if this is right because it seems a little short. Can someone let me know if this is the right approach? Thanks, Chad.

Edit: Maybe I have to use Bolzano-Weierstrass and say that $\displaystyle x_n$ has a convergent sub sequence that tends to a as x approaches infinity, not really sure that makes a difference though since we already have a sequence that converges to a.