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?

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- Feb 4th 2011, 05:04 AMmaximus101Analysis question
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? - Feb 4th 2011, 05:23 AMPlato
If $\displaystyle \varepsilon > 0$ and $\displaystyle 0<|x-c|<\varepsilon$ then one of these is true, $\displaystyle x\in (c-\varepsilon ,c)\text{ or }x\in (c,c+\varepsilon)$.

The first is involved with the limit from the left the other the limit from the right. - Feb 4th 2011, 05:34 AMMath Major
Let $\displaystyle \epsilon > 0 $ be given. By assumption, we can find a $\displaystyle \delta_1, \delta_2$ such that for $\displaystyle x \in (c - \delta_1, c) $ or $\displaystyle x \in (c, c + \delta_2$, we have that $\displaystyle |f(x) - 1| < \epsilon $.

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