TITLE: Existence of divergent subsequence
QUESTION [0 upvotes]: Let $f: [a,b) \rightarrow \mathbb{R}$ ($a<b$ are finite). Suppose $\lim _{x \rightarrow b^-} f(x)=+\infty$. I'd like to find a sequence $b_n \in [a,b)$ converging to $b$ such that $\lim _{n} f(b_n)=+\infty$. Is it possible?
I tried to derive this from the $\epsilon - \delta$ definition of limit. I sure can get a sequence $c_n \in [a,b)$ such that $f(c_n)$ goes to $+\infty$, but I cannot find a way to ensure the convergence of $c_n$ to $b$.

REPLY [1 votes]: Yes. There is some $\delta_{1}>0$ such that $f(x)>1$ for all $b-\delta_{1}<x<b$. Take a $b_{1}$ such that $b-\delta_{1}<b_{1}<b$, so $f(b_{1})>1$. There is some $\delta_{2}>0$ such that $f(x)>2$ for all $b-\delta_{2}<x<b$. Take a $b_{2}$ such that $b-\delta_{2}<b_{2}<b$ and $b_{2}>b_{1}$, for example, we must have $\max\{b-\delta_{2},b_{1}\}<b$, then simply choose $b_{2}=(1/2)(\max\{b-\delta_{2},b_{1}\}+b)$, then $f(b_{2})>2$. 
Proceed in this way and you got that.
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