Show $\frac{1}{x^{2}}$ is not uniformly continuous on (0,5]. Show g(x) = $\frac{1}{x^{2}}$ is not uniformly continuous on (0,1.5].
attempt
g(x) is not uniformly continuous if 
$$\exists \epsilon > 0 : \forall m > 0 \exists x,y \epsilon (0,1.5]  |x-y| < m, |g(x)-g(y)| \geq \epsilon$$
$ |g(x)-g(y)| = |\frac{1}{x^{2}} - \frac{1}{y^{2}}| = |\frac{y^{2}-x^{2}}{x^{2}y^{2}}|$
$$$$
$=\frac{|x^{2}-y^{2}|}{x^{2}y^{2}} = \frac{|x-y||x+y|}{x^{2}y^{2}} = \frac{|x-y||x-y+2y|}{x^{2}y^{2}} \geq \frac{|x-y|(|x-y|-2|y|)}{x^{2}y^{2}} = \frac{|x-y|(1-2|y|)}{x^{2}y^{2}}$ if m = 1
$$$$
and if $\frac{|x-y|(1-2|y|)}{x^{2}y^{2}}$ is to be made less than $\epsilon$ then $|x-y|) < \frac{x^{2}y^{2}\epsilon}{(1-2|y|}$ and since $x^{2}y^{2}$ is smaller and smaller for x and y close to 0, no single number m workds for all x and y and so g(x) = $\frac{1}{x^{2}}$ is not uniformly continuous on (0,1.5].
 A: Let $\epsilon =1$, assume to get contradiction that $g(x)=\frac{1}{x^2}$ is uniformly continuous on $(0,1.5]$. By the assumption exists $\delta > 0$ s.t. $\forall x,y\in (0,1.5],\mid x-y\mid<\delta \Rightarrow \mid g(x)-g(y) \mid <\epsilon$. Let us observe the following series, $x_n=\frac{1}{n}$ and $y_n=\frac{1}{2n}$. It is easy to see that $\mid x_n-y_n\mid=\mid \frac{1}{n}-\frac{1}{2n}\mid=\mid \frac{1}{2n}\mid \xrightarrow{n\rightarrow \infty} 0$ thus for a large enough $n$ we have $\mid x_n-y_n\mid<\delta$. But we also have $\mid g(x_n)-g(y_n)\mid=\mid \frac{1}{(\frac{1}{n})^2}-\frac{1}{(\frac{1}{2n})^2}\mid=\mid n^2-4n^2\mid=3n^2\xrightarrow{n\rightarrow \infty}\infty$. For large enough $n$ we have $\mid x_n-y_n\mid<\delta$ and $\mid g(x_n)-g(y_n)\mid >1=\epsilon$
A: Perhaps I'm greatly mistaken, but I believe that it IS uniformly continuous on (0,1.5], Since f(x) is (in the usual sense) continuous on (0,1.5], and (0,1.5] is compact, then f(x) is uniformly continuous on (0,1.5].  (This is a basic theorem regarding uniform continuity and compactness)
