# Thread: Most likely an easy continuity question

1. ## Most likely an easy continuity question

Hi everyone, this is probably an easy question but I'm having trouble on the wording of the proof.

Let f be continuous at x=c and f(c) > 1

Show that there exists an r > 0 such that $\displaystyle \forall x \in B(c,r) \bigcap D : f(x) > 1$

Like I said, probably easy, but I don't know how to show it on paper.

2. Originally Posted by HeirToPendragon
Hi everyone, this is probably an easy question but I'm having trouble on the wording of the proof.

Let f be continuous at x=c and f(c) > 1

Show that there exists an r > 0 such that $\displaystyle \forall x \in B(c,r) \bigcap D : f(x) > 1$

Like I said, probably easy, but I don't know how to show it on paper.
Use the definition of "continuity" taking $\displaystyle \epsilon= \frac{f(c)}{2}$.

3. Originally Posted by HallsofIvy
Use the definition of "continuity" taking $\displaystyle \epsilon= \frac{f(c)}{2}$.
It may be that $\displaystyle \frac{f(c)}{2} < 1$

To fix it consider: $\displaystyle \varepsilon = \frac{{f(c) - 1}} {2}\, \Rightarrow \,\left| {f(x) - f(c)} \right| < \frac{{f(c) - 1}} {2}\,\, \Rightarrow \,1 < \frac{{f(c) + 1}} {2} < f(x)\,$.

4. Originally Posted by Plato
It may be that $\displaystyle \frac{f(c)}{2} < 1$

To fix it consider: $\displaystyle \varepsilon = \frac{{f(c) - 1}} {2}\, \Rightarrow \,\left| {f(x) - f(c)} \right| < \frac{{f(c) - 1}} {2}\,\, \Rightarrow \,1 < \frac{{f(c) + 1}} {2} < f(x)\,$.
Yep, that works out perfectly. Thanks.

5. Thanks, Plato. I was thinking of the more common problem, f(x)> 0, and didn't pay attention to the "> 1" part.