not uniformly continuous?

Oct 2009
15
0
How do you show f(x)=1/x is not uniformly continuous?
 

chisigma

MHF Hall of Honor
Mar 2009
2,162
994
near Piacenza (Italy)
Given \(\displaystyle \varepsilon >0\) and \(\displaystyle M>0\), it exists at least one value \(\displaystyle x_{0}\) for which is \(\displaystyle |f(x_{0} + \varepsilon) - f(x_{0})|>M\) or \(\displaystyle |f(x_{0} - \varepsilon) - f(x_{0})|>M\)...

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
How do you show f(x)=1/x is not uniformly continuous?
Answer this question: "Is \(\displaystyle f(x)=x\) greater than zero"?
 
Feb 2010
100
27
Lebanon - Beirut
Take two sequences \(\displaystyle x_n = \frac{1}{n}\) and \(\displaystyle y_n = \frac{1}{n+1}\). Note that \(\displaystyle \lim_{n\rightarrow \infty}x_n = \lim_{n\rightarrow \infty}y_n = 0\).
But: \(\displaystyle f(y_n)-f(x_n) = \frac{1}{\frac{1}{n+1}} - \frac{1}{\frac{1}{n}} = 1\). This means that you can take two terms: one from \(\displaystyle {x_n}\) and the other from \(\displaystyle {y_n}\) such that both terms are near each other (since both sequences approach zero) but the absolute difference between their function values is 1. Hence, this contradicts the property of uniform continuity, i.e. f is not uniformly continuous.

hope this helps (Cool)
 
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