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?