Prove that $f=1/\sqrt{x}$ is continuous on the interval $(0,1]$, but not uniformly continuous. Prove that $f(x)=1/\sqrt{x}$ is continuous on the interval $(0,1]$, but not uniformly continuous.
I believe it follows that $f(x)$ is not uniformly continuous because $f(x)$ is not continuous on the bounded interval including zero. I tried to set up an $\epsilon-\delta$ proof of the continuity on $(0,1]$ with $\delta=\frac{p}{(2\epsilon+1)^2}$ where $p$ is any point in the interval, but I feel like I am making a leap to say that by that $\delta$ we get $|\frac{\sqrt p-\sqrt x}{\sqrt p\sqrt x}|\lt\epsilon$
 A: Recall that $f$ is uniformly continuous if and only if for each $\varepsilon>0$, there exists $\delta>0$ such that
$$\sup\{|f(x)-f(y)| : |x-y|<\delta\}<\varepsilon.$$
Consider the sequence $x_n=\frac1{n^2}$. Then for any $N$,
$$
|x_N-x_{N+m}| = \frac1{N^2}-\frac1{(N+m)^2} \stackrel{m\to\infty}\longrightarrow\frac1{N^2},
$$
and
 $$|f(x_N)-f(x_{N+m})|=\left|\frac1N-\frac1{N+m} \right|=\frac{m}{N(N+m)}. $$
Choose $\varepsilon=\frac12$. Then for any $\delta>0$, we may choose $N$ such that $\frac1{N^2}<\delta$, but
$$f(x_N)-f(x_{N+m}) =  \frac{m}{N(N+m)}> \frac m{\delta+m}\stackrel{m\to\infty}\longrightarrow 1,$$
so for $m$ sufficiently large, $$|f(x_N)-f(x_{N+m})|>\frac12.$$ It follows that $f$ is not uniformly continuous
A: Here we can use the following sequential criterion:

If $f$ is not uniformly continuous on a given domain $D$, then we can find two sequences $(x_n)$ and $(y_n)$ in $D$ such that $\lim_{n\to\infty}(x_n-y_n)=0$ and $|f(x_n)-f(y_n)|\ge\epsilon_0$ for some fixed positive $\epsilon_0.$ 

Simply choose $x_n=\dfrac{1}{n^2}$ and $y_n=\dfrac{1}{4n^2}$ for all $n\in\mathbb{N}.$
A: Let $a<b$ and let $J$ be any of $[a,b],[a,b),(a,b], (a,b).$ Let $f:J\to \Bbb R$ be uniformly continuous. Then $f$ is bounded on $J.$
Proof:  Take $\epsilon>0$ such that $\forall x,y \in J\,(|x-y|<\epsilon\implies |f(x)-f(y)|<1).$ Take $n\in \Bbb N$ such that $1/n<\epsilon$ and $1/2n<b-a.$
For $j\in \Bbb N$ let $x_j=a+j/2n.$ We have $x_1\in J.$
Take the  unique $k\in \Bbb N$ such that $x_k\le b< x_{k+1}.$
Now for any $y\in J$ we have: 
(i).  $y< x_2\implies |y-x_1|\le 1/2n\implies |y-x_1|<1/n\implies |f(y)-f(x_1)|<1.$
(ii). If $y\ge x_2$ then there is a unique $i\in \Bbb N$ with $2\le i\le k$ and $x_i\le y< x_{i+1}$. We have $$(\bullet)\quad |f(x_1)-f(y)|\le \left(\sum_{j=1}^{i-1}|f(x_j)-f(x_{j+1})|\right) +|f(x_i)-f(y)|.$$ Now each $|x_j-x_{j+1}|=1/2n<1/n$ so each $|f(x_j)-f(x_{j+1})|<1.$ And $|x_i-y|<1/2n<1/n$ so $|f(x_i)-f(y)|<1.$ So the RHS value  in $(\bullet)$ is less than $(i-1)+1=i\le k.$
So if $x_2\le y\in J$ then $|f(x_1)-f(y)|<k.$
(iii). From (i) and (ii)  we have $\forall y\in J\, (f(x_1)-k<f(y)<f(x_1)+k).$ QED.
Now $f(x)=1/\sqrt x\,$ is unbounded on $(0,1]$ so it cannot be uniformly continuous on $(0,1].$
