# The image of an $\epsilon$-neighborhood of a function with a nonzero limit.

The following is from Gaughan's Introduction to Analysis:

Suppose $$g:D\rightarrow \mathbb{R}$$ with $$x_0$$ and accumulation point of $$D$$ and $$g(x)\neq 0$$ for all $$x\in D$$. Further, assume that g has a limit at $$x_0$$, and $$lim_{x\rightarrow x_0}f(x)\neq 0$$. Prove that There is a positive $$\epsilon_0$$ and positive $$M$$ such that for all $$x\in (x_0-\epsilon_0, x_0+\epsilon_0)\cap D$$, $$|f(x)|\geq M$$.

Here is what I have tried so far: Since $$f(x)$$ has a limit, L, at $$x_0$$, for every $$\epsilon>0$$ there exists some positive $$\delta$$ such that whenever $$x\in D$$ and $$0<|x-x_0|<\delta$$, $$|f(x)-L|<\epsilon$$. If we set $$\epsilon_0 = \delta$$, then we have $$-\epsilon_0<0<|x-x_0|<\epsilon_0$$, so $$x_0-\epsilon, and thus $$x\in (x_0-\epsilon_0,x_0+\epsilon_0)\cap D$$. We know that under these conditions, for every $$\epsilon$$ we choose, $$|f(x)-L|<\epsilon$$. This is where I'm stuck. It seems like there should be some $$\epsilon$$ that we can choose so that supposing $$|f(x)| gives us a contradiction, or something like that.

Thanks!

• Did you mean $(x_0-\varepsilon_0,x_0+\varepsilon_0)$ instead of $(x-\varepsilon_0,x+\varepsilon_0)$? – José Carlos Santos Oct 22 '18 at 21:53
• Yep, thank you! – Wyatt Kuehster Oct 23 '18 at 1:55

Take $$\epsilon =\frac {|L|} 2$$ and use $$|f(x)| \geq |L| -|f(x)-L|>\frac {|L|} 2$$.
• And then we'd get M=L-|f(x)-L|>0 and $\epsilon = \delta>0$, so I get that (I think). Doesn't that suppose that $L>0$? If $L<0$, can we take $\epsilon=\frac{-L}{2}$ and let $M=-L-|f(x)-L|$? – Wyatt Kuehster Oct 23 '18 at 2:36