# not uniformly continuous?

#### Kiwili49

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

#### chisigma

MHF Hall of Honor
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
How do you show f(x)=1/x is not uniformly continuous?
Answer this question: "Is $$\displaystyle f(x)=x$$ greater than zero"?

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.