If $f$ is continuous on $(0,1)$ and $\lim_{x \to 0^+}f(x) = +\infty$, show that $f$ is not uniformly continuous 
If $f$ is continuous on $(0,1)$ and $\lim_\limits{x \to 0^+}f(x) = +\infty$, show that $f$ is not uniformly continuous.

Edit: There is a similar question that has been answered, but this one discusses the reasoning behind certain details in more depth.
Attempt
Suppose $f$ is uniformly continuous. Then $\forall\epsilon > 0$ there exists $\delta > 0$ s.t for all $x,y \in S \subset \mathbb{R}$, if $\lvert x-y\rvert < \delta$ then $\lvert f(x) - f(y)\rvert < \epsilon$.
But $\lim\limits_{x \to 0^+}f(x) = +\infty \implies \lvert f(x) -f(y)\rvert = +\infty > \epsilon$.
Comment: I know that this is not formally right due to how I'm using infinity, but I can't figure out how to formalize what I'm trying to say. I understand what is going on with the function and why it can't be uniformly continuous, but expressing it is becoming difficult.
 A: Let's suppose it were uniformly continuous. Pick $\varepsilon > 0$ and the corresponding $\delta > 0$ as in the definition. Choose also an $x$ with $0 < x < \delta$. By uniform continuity, for any $y$ with $0 < y < x$, $|f(x) - f(y)| < \epsilon$. But then if we take the limit as $y$ approaches $0$ from the right, then $$\lim_{y\to0^+}|f(y)| \le |f(x)| + \lim_{y\to0^+}|f(y) - f(x)| < |f(x)| + \epsilon.$$
This contradicts the fact that $\lim_{x\to0^+}f(x) = +\infty$, so $f$ cannot be uniformly continuous.
A: Suppose that $f$ is uniformly continuous. Then there is a $\delta>0$ such that$$\lvert x-y\rvert<\delta\implies\bigl\lvert f(x)-f(y)\bigr\rvert<1.$$Now, take $N\in\mathbb N$ such that $N>\frac1\delta$. If $x\in\left(0,\frac12\right)$, then there are numbers$$x_0(=x)<x_1<x_2<\cdots<x_n\left(=\frac12\right)$$with $n\leqslant N$ and $x_k-x_{k-1}<\delta$. But then$$\left\lvert f(x)-f\left(\frac12\right)\right\rvert=\bigl\lvert f(x_0)-f(x_n)\bigr\rvert\leqslant n\leqslant N.$$So, you cannot have $\lim_{x\to0^+}f(x)=+\infty$.
A: Assume f uniformly continuous on (0,1).
Fix $\epsilon >0$; There exists a $\delta >0$ s,t.
$|x-y| <\delta$ implies $|f(x)-f(y)|> \epsilon$ , $x,y \in (0,1)$.
For this $\delta$ there exists a $m$ with $1/m <\delta$ (Archimedean principle).
Set $y=1/m$(fixed).
Choose $n >m$, then $1/m -1/n <\delta$.
Set $x_n=1/n$.
Then 
$|x_n-y| <\delta$ implies $|f(x_n)-f(y)| < \epsilon.$
$f(x_n)< \epsilon +f(y)$, i.e $f(x_n)$ is bounded above,
a contradiction since  
for $\lim_{n \rightarrow \infty} x_n =0$ we have$\lim_{n \rightarrow \infty}f(x_n) =\infty.$
