1/x and continuity

**I-Think** · January 19th 2011, 10:32 PM

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 $(x_n-p_n)\rightarrow{0}$

$f(x_n)-f(p_n)=\frac{1}{x_n}-\frac{1}{p_n}=\frac{p_n-x_n}{x_np_n}<br />$
But this would seem to converge to 0, by assumption.

Can someone point out the flaw in my reasoning?
Please and thank you

**Drexel28** · January 19th 2011, 10:39 PM

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 $(x_n-p_n)\rightarrow{0}$

$f(x_n)-f(p_n)=\frac{1}{x_n}-\frac{1}{p_n}=\frac{p_n-x_n}{x_np_n}<br />$
But this would seem to converge to 0, by assumption.

Can someone point out the flaw in my reasoning?
Please and thank you

Consider that $\displaystyle \frac{1}{n}-\frac{1}{n^2}\to 0$ yet $\displaystyle n-n^2\not \to 0$ .

**I-Think** · January 19th 2011, 11:01 PM

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 $0$ ?

**Drexel28** · January 19th 2011, 11:13 PM

Originally Posted by I-Think

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 $0$ ?

Yes, I don't see why not. I don't see anything that doesn't serve to perfectly dissatisfy the hypotheses of uniform continuity.

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