Proving $f(x)=1/x$ on $(0,1 )$ is not uniformly continuous My questions are about the reasoning made in the note http://folk.uib.no/st00895/MAT112-V12/unif-kont.pdf (which is in Norwegian). 
To prove that $f(x)=\frac{1}{x}$ is not uniformly continuous, the authors use the following "result" (Sats 2.12 in the note, which I translate below):
Result 2.12: If for every $h>0$ we have that $|f(x+h)-f(x)|$ is unbounded on $I$, then $f$ is not uniformly continuous on $I$.
Proof: The result follows directly from the definition of uniform continuity. 
In Example 2.14 (Eksempel 2.14), the authors look at
$$|f(x+h)-f(x)|=\left|\frac{1}{x+h}-\frac{1}{x}\right|=\left|\frac{h}{x(x+h)}\right|.$$
They then claim that the above quantity is not bounded for any $h>0$, since 
$$\lim_{x\rightarrow 0}\left|\frac{h}{x(x+h)}\right|=\infty.$$
Question 1: Is it not possible to choose $h=x^2$, thereby obtaining 
$$\frac{h}{x(x+h)}=\frac{1}{\frac{x}{h}(x+h)}=\frac{1}{x(1/x+1)}=\frac{1}{x(1/x+1)}=\frac{1}{1+x}\rightarrow1 \text{ as }x\rightarrow 0.$$
Therefore, |f(x+h)-f(x)| is not unbounded, so we cannot use the result "Result 2.12". Is my argumentation correct?? Is it OK to choose $h$ like I have done? 
Question 2: However, it seems correct to me that my argumentation is all you need to prove that  $f(x)=\frac{1}{x}$ is not uniformly continuous on (0,1). Here $x_1=x,x_2=x+x^2$, so that $|x_1-x_2|$ can be made arbitrarily small. However $|f(x_1)-f(x_2)|=1$, which stays the same regardless of how small we make $|x_1-x_2|$. Is this correct?
 A: To use the result you cite, $h$ must be a constant. It cannot depend on $x$. Your argument is that no matter how small $|x_1-x_2|$ is, $$\left|\frac{1}{x}-\frac{1}{x+x^2}\right|=\left|\frac{x^2}{x^2(x+1)}\right|=\left|\frac{1}{1+x}\right|$$ is close to $1$ as $x$ is close to zero. This will work, but you need to use the $\epsilon$-$\delta$ formalism to prove it (or use the result).
A: you can use another mehtod to show f(x)=1/x is not uniformly continuous on (0,1)
let define :${ x }_{ n }=\frac { 1 }{ n+1 } ,{ y }_{ n }=\frac { 1 }{ n+1+\varepsilon  } $
$$\left| { x }_{ n }-{ y }_{ n } \right| =\frac { \varepsilon  }{ \left( n+1 \right) \left( n+1+\varepsilon  \right)  } \rightarrow 0  $$ $when$  $n\rightarrow \infty $
however,$$\\ \\ \quad \\ \left| f\left( { x }_{ n } \right) -f\left( { y }_{ n } \right)  \right| =\left| n+1-n-1-\varepsilon  \right| =\varepsilon .\forall \varepsilon >0\\ $$ which shows f(x) is not not uniformly continuous
A: Here's another proof that might make it more obvious.
All we have to show is that for any $\varepsilon > 0$ and any $\delta > 0$, we can find points $p,q \in (0,1)$ such that $|p-q| < \delta$ and $\left|\frac{1}{p}-\frac{1}{q} \right| > \varepsilon$.
Just take $q = \frac{p}{2}$: now the conditions are $\frac{p}{2} < \delta$ and $\varepsilon < \frac{1}{p}$, and both will be satisfied when $0 < p < min\left\{ 2\delta, \frac{1}{\varepsilon}, 1 \right\}$.
