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I've been given an interesting proof question and have attempted it myself. Although I roughly understand how to go about the proof, I have one critical doubt, which I will expound upon later.

The following is what needs to be proven:

Given:

$\lim_{x\to a^-} f(x)= \infty $

$\lim_{x\to a^+} f(x)= -\infty $

use only the precise definitions of the limit and infinite limit to prove that

$\lim_{x\to a} {1\over f(x)}= 0 $


My attempt

Suppose that:

1) For all $M>0$, there exists $\delta_1 >0$ such that:

$0<a-x<\delta_1\implies f(x)>M$

2)For all $N<0$, there exists $\delta_2 >0$ such that:

$0<x-a<\delta_2\implies f(x)<N$

What we need to show is:

For all $\epsilon>0$, there exists a $\delta>0$ so that

$0<|x-a|<\delta\implies |{1\over f(x)}|<\epsilon$

Consider RHS:

$|{1\over f(x)}|<\epsilon$

$ \implies |f(x)|>{1\over \epsilon}$

Choose $\delta=min(\delta_1, \delta_2)$

Then

$0<|x-a|<\delta\implies f(x)>M \land f(x)<N$

i.e. $0<|x-a|<\delta\implies M<f(x)<N$

Choose $M={1\over \epsilon}$

Therefore, from our assumptions, we have shown there exists a $\delta$ such that

$0<|x-a|<\delta\implies |{1\over f(x)}|<\epsilon$


I would like your help in assessing whether this proof is valid. I spot one issue in it myself: how can the following be?

"$M<f(x)<N$"

We know $M>0$ and $N<0$. I wonder what logical missteps I took to reach this conclusion.

Thank you.

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You have got the right idea and there is slight problem at the end. You start with the following statements which are correct:

  1. For every $M > 0$ there is a $\delta_{1} > 0$ such that $f(x) > M$ whenever $0 < a - x < \delta_{1}$.
  2. For every $N < 0$ there is a $\delta_{2} > 0$ such that $f(x) < N$ whenever $0 < x - a < \delta_{2}$.

Now we have to use these two statements above to prove the truth of the following statement:

For any $\epsilon > 0$ there is a $\delta > 0$ such that $|1/f(x)| < \epsilon$ whenever $0 < |x - a| < \delta$.

We observe that $|1/f(x)| < \epsilon$ means that $|f(x)| > 1/\epsilon$ and this is equivalent to "$f(x) < -1/\epsilon$ or $f(x) > 1/\epsilon$". Choose $M = 1/\epsilon, N = -1/\epsilon$ and get $\delta_{1}, \delta_{2}$ based on $M, N$ (this is possible because of the two statements mentioned in the beginning. Let $\delta = \min(\delta_{1}, \delta_{2})$ and this will ensure that if $0 < |x - a| < \delta$ then either $f(x) < -1/\epsilon$ or $f(x) > 1/\epsilon$ and therefore $|1/f(x)| < \epsilon$.

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  • $\begingroup$ Thank you. The handling of $1\over f(x)$ at the end was what I was most confused about! The 'or'-ing of the statement naturally comes from the modulus - this I did not realize. $\endgroup$ – Sar T1 Sep 25 '16 at 17:08
  • $\begingroup$ @SarT1: you are right. I am glad that my answer helped you to identify your mistake. $\endgroup$ – Paramanand Singh Sep 25 '16 at 19:26

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