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. limxcf(x)=0) how to prove that this implies: limxc1f(x)=.




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


Answer



This is problematic, even if you consider 1/|f(x)| instead of 1/f(x). For example, let f(x)={xsin(1/x)x00otherwise.

This is everywhere defined and continuous on R, and limx0f(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 limx01|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 ER and f:ER. Let F={xE:f(x)0}. Suppose further that cR is a limit point of both E and F, and that for all ϵ>0 there is some δ>0 such that |f(x)|<ϵ whenever xE with 0<|xc|<δ. Then for all M, there exists δ>0 such that 1/|f(x)|>M whenever xF with 0<|xc|<δ.


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