1. ## 1/x and continuity

Prove $f(x)=\frac{1}{x}$ is not uniformly continuous over $(0,1)$

There are multiple ways to prove this, but I'm having trouble with a particular one

It is known that a $f:I\rightarrow{R}$ is uniformly continuous if and only if when $x_n,p_n\in{I}$ and $(x_n-p_n)\rightarrow{x}$, then $(f(x_n)-f(p_n))\rightarrow{0}$

I'm trying to use this result to prove the question
So if $(x_n-p_n)\rightarrow{0}$

$f(x_n)-f(p_n)=\frac{1}{x_n}-\frac{1}{p_n}=\frac{p_n-x_n}{x_np_n}
$

But this would seem to converge to 0, by assumption.

Can someone point out the flaw in my reasoning?

2. Originally Posted by I-Think
Prove $f(x)=\frac{1}{x}$ is not uniformly continuous over $(0,1)$

There are multiple ways to prove this, but I'm having trouble with a particular one

It is known that a $f:I\rightarrow{R}$ is uniformly continuous if and only if when $x_n,p_n\in{I}$ and $(x_n-p_n)\rightarrow{x}$, then $(f(x_n)-f(p_n))\rightarrow{0}$

I'm trying to use this result to prove the question
So if $(x_n-p_n)\rightarrow{0}$

$f(x_n)-f(p_n)=\frac{1}{x_n}-\frac{1}{p_n}=\frac{p_n-x_n}{x_np_n}
$

But this would seem to converge to 0, by assumption.

Can someone point out the flaw in my reasoning?
Consider that $\displaystyle \frac{1}{n}-\frac{1}{n^2}\to 0$ yet $\displaystyle n-n^2\not \to 0$.
3. Just thought of this, given that the theorem works only if for any $x_n, p_n$ in the interval $I$, and
Would it fail in this case as if we have $(x_n)\rightarrow{0}$ and $(p_n)\rightarrow{0}$,we have $(x_n-p_n)\rightarrow{0}$ but $(f(x_n)-f(p_n))$ does not converge to $0$?
Just thought of this, given that the theorem works only if for any $x_n, p_n$ in the interval $I$, and
Would it fail in this case as if we have $(x_n)\rightarrow{0}$ and $(p_n)\rightarrow{0}$,we have $(x_n-p_n)\rightarrow{0}$ but $(f(x_n)-f(p_n))$ does not converge to $0$?