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 $x_0-\epsilon<x<x_0+\epsilon$, 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 $|f(x)|<M$ gives us a contradiction, or something like that.


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

Take $\epsilon =\frac {|L|} 2$ and use $|f(x)| \geq |L| -|f(x)-L|>\frac {|L|} 2$.

  • $\begingroup$ 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 $M=-L-|f(x)-L|$? $\endgroup$ – Wyatt Kuehster Oct 23 '18 at 2:36
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    $\begingroup$ @WyattKuehster I have corrected my answer. $\endgroup$ – Kavi Rama Murthy Oct 23 '18 at 5:05

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