Analysis question

**maximus101** · February 4th 2011, 05:04 AM

suppose that f : (a,b)\{c} ----> real numbers is a function such that

lim (x--->c+) {f(x)} and lim (x--->c-) {f(x)} both exist and are equal to a common value l.

how can we prove that lim (x--->c) {f(x)} exists and that it equals l?

**Plato** · February 4th 2011, 05:23 AM

If $\varepsilon > 0$ and $0<|x-c|<\varepsilon$ then one of these is true, $x\in (c-\varepsilon ,c)\text{ or }x\in (c,c+\varepsilon)$ .
The first is involved with the limit from the left the other the limit from the right.

**Math Major** · February 4th 2011, 05:34 AM

Let $\epsilon > 0$ be given. By assumption, we can find a $\delta_1, \delta_2$ such that for $x \in (c - \delta_1, c)$ or $x \in (c, c + \delta_2$ , we have that $|f(x) - 1| < \epsilon$ .

Set $\delta = min(\delta_1, \delta_2)$ . Then for $x \in (c - \delta, c+\delta), x \ne c$ , it follows that $|f(x) - 1| < \epsilon$ .

Thread: Analysis question

LinkBack

Thread Tools

Display

Analysis question

Similar Math Help Forum Discussions

Analysis Question

Analysis question no.2

Analysis Question

analysis question

analysis question

Search Tags