Sunday, 22 January 2017

limits - limxtocf(x)=0Rightarrowlimxtoc1overf(x)=infty



If f is defined as a function of real variables to real values, and ccl(Domain) as its limit value (i.e. lim) 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].


Answer



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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