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If $f$ is defined as a function of real variables to real values, and $c \in cl(Domain)$ as its limit value (i.e. $\lim_{x \to c} {f(x)} = 0 $) how to prove that this implies: $\lim_{x \to c} {1 \over f(x)} = \infty$.

It seems logical that the values will be always bigger, but when tried to construct a contradiction using the y-creterion I stuck at: $\exists \epsilon > 0: f(x)>0 \forall x \in [c-\epsilon,c+\epsilon]$.

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It's not true unless you consider $1/|f(x)|$ instead of $1/f(x)$. For instance take $f(x) = x$ and $c=0$. –  Antonio Vargas May 7 '13 at 23:24
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2 Answers

up vote 3 down vote accepted

This is problematic, even if you consider $1/|f(x)|$ instead of $1/f(x)$. For example, let $$f(x)=\begin{cases}x\sin(1/x) & x\ne0\\ 0 & \text{otherwise.}\end{cases}$$ This is everywhere defined and continuous on $\Bbb R$, and $$\lim_{x\to 0}f(x)=0,$$ but since there is no $x$-interval around $0$ on which $1/|f(x)|$ is defined, then it is problematic to talk about $$\lim_{x\to0}\frac1{|f(x)|}.$$ It's even more problematic to talk about it if we were to let $f$ be the constant zero function.

We must make some extra assumptions to take care of your problems. In particular, you need to show the following:

Suppose that $E\subseteq\Bbb R$ and $f:E\to\Bbb R.$ Let $F=\{x\in E:f(x)\ne0\}.$ Suppose further that $c\in\Bbb R$ is a limit point of both $E$ and $F,$ and that for all $\epsilon>0$ there is some $\delta>0$ such that $|f(x)|<\epsilon$ whenever $x\in E$ with $0<|x-c|<\delta$. Then for all $M,$ there exists $\delta>0$ such that $1/|f(x)|>M$ whenever $x\in F$ with $0<|x-c|<\delta.$

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Good point. I guess the limit should be taken in the domain of $1/f$. –  Antonio Vargas May 7 '13 at 23:45
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Let me just assume that $c = 0$. And as pointed out in the comment above, you have to be careful. Take the following just as an outline of a general approach. As pointed out in the excellent answer by Cameron, this doesn't always work. You have to make some assumptions. But maybe it can be helpful?

But, I assume that what want to prove that given $N$ there is an $\delta>0$ such that $$ \lvert x\rvert < \delta \Rightarrow \lvert f(x)^{-1}\rvert > N. $$ So Let $N$ be given. Let $\epsilon = \frac{1}{N}$. Then there is a $\delta >0$ such that $\lvert f(x)\rvert < \epsilon = N^{-1}$. That is you get exactly what you want because $$\begin{align} \lvert f(x)\rvert &< \epsilon = N^{-1} \quad \Rightarrow \\ \lvert f(x)\rvert^{-1} &> N. \end{align} $$

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This doesn't work (at least, not quite so simply). See my answer. –  Cameron Buie May 7 '13 at 23:43
    
@CameronBuie: You are right! I updated my answer. –  Thomas May 7 '13 at 23:56
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