0
$\begingroup$

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.

Thanks!

$\endgroup$
  • 1
    $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @WyattKuehster I have corrected my answer. $\endgroup$ – Kavi Rama Murthy Oct 23 '18 at 5:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.