Show that $f(x)=\ln(x)$ is not uniformly continuous on $(0,\infty)$ I'm trying to show that $f(x)=\ln(x)$ is not uniformly continuous on the interval $(0,\infty).$
This is what I have so far:
Let $\epsilon=1.$
Choose $\delta=$
if $x,y\in(0,\infty)$ with $|y-x|<\delta$ then $|f(y)-f(x)|=|\ln\left(\frac{y}{x}\right)|$
I'm stuck at this point though, are there any well known inequalities I can use here?
 A: The Cauchy sequence $x_n = 1/n$ gets mapped, by $\ln$, to a divergent (hence non-Cauchy) sequence.
A: Working with $\epsilon$ and $\delta$ quickly becomes tedious and annoying, it is thus better to learn more convenient and powerful techniques. Remember that $f$ is uniformly continuous on $S$ if and only if for every sequences $(x_n), (y_n) \subseteq S$ with $d(x_n, y_n) \to 0$ we have that $d(f(x_n), f(y_n)) \to 0$ (with $d$ denoting the distances in the domain the definition and the range of $f$).
In our case, we suspect that $\ln$ fails to be uniformly continuous towards of $0$. Choose, therefore, $x_n = \textrm e^{-n}$ and $y_n = \textrm e^{-n + 1}$. Notice that $|x_n - y_n| \to |0 - 0| = 0$, but $$| \ln x_n - \ln y_n | = | \ln \textrm e^{-n} - \ln \textrm e^{-n + 1} | = | -n - (-n + 1)| = 1 \not\to 0 ,$$
which shows that $\ln$ is not uniformly continuous on $(0,\infty)$. (It is, though, on every interval of the form $[a,b)$ with $a > 0$ and $b$ possibly infinite.)
A: Let $\epsilon=1$ be like you've chosen. Then for all $\delta>0$, just pick $y=\frac{3}{4}\delta>0$ and $x=\frac{1}{4}\delta>0$. This works because
$$
|y-x|=\frac{1}{2}\delta<\delta\quad\text{but}\quad|\log(y)-\log(x)|=\log(3)>1=\epsilon.
$$
A: General fact: No unbounded function on a bounded subset of $\mathbb R$ can be uniformly continuous. (In the problem at hand, we have $\ln x $ unbounded on $(0,1),$ so it can't be uniformly continuous there.)
A: \begin{align}
    \left| \ln \dfrac{y}{x} \right| < 1 \\
   -1 < \ln \frac{y}{x} < 1 \\
    \dfrac 1e < \dfrac{y}{x} < e \\
    \dfrac xe < y < ex \\
\end{align}
work to find numbers to use in proof

It will suffice to show that there exists $0 < x < y$ such that $y-x < \delta$ and yet $y > ex$. So lets $y = x + \dfrac{\delta}{2}$ and see what we can do with $y > ex$.
\begin{align}
    x + \dfrac{\delta}{2} &> ex \\
    (e - 1)x &< \dfrac{\delta}{2} \\
    x &< \dfrac{\delta}{2(e-1)}
\end{align}
Solution

For any $\delta > 0$, let $x = \dfrac{\delta}{4(e-1)}$ and let 
$y = x + \dfrac{\delta}{2}$
Then $|y - x| = y - x = \dfrac{\delta}{2} < \delta$
and
\begin{align}
    \left| \ln \dfrac{y}{x} \right|
      &=\left| \ln \dfrac{\dfrac{\delta}{4(e-1)} + \dfrac{\delta}{2}}
                         {\dfrac{\delta}{4(e-1)}} \right| \\
      &=\left| \ln (1+ 2(e-1)) \right|\\
      &=\left| \ln (e + (e - 1)) \right|\\
      &> 1
\end{align}
It follows that $y = \ln(x)$ is not uniformly continuous on $(0, \infty)$.
A: Well, it all depends $\ldots$

Proposition: Let $(\mathbb{R}_{>0},d_A)$ and $(\mathbb{R},d_B)$ metric spaces with $d_A(x,y) = | \ln(x) - \ln(y) |$ and $d_B(x,y) = | x - y |$. Then
  $$
f : \mathbb{R}_{>0} \to \mathbb{R}, \, x \mapsto \ln(x)
$$ is uniformly continuous.
Proof: Let $\epsilon > 0$ and choose $\delta := \epsilon$. Then for all $x,y \in \mathbb{R}_{>0}$ we have
  $$
d_A(x,y) < \delta \quad\Longrightarrow\quad d_B(f(x),f(y)) < \epsilon.
$$
  $$\tag*{$\square$}$$

Notice that $(\mathbb{R}_{>0},d_A) \simeq (\mathbb{R},d_B)$. Since every isometry between any two metric spaces is uniformly continuous, so is the isometry $f := \ln$ between $(\mathbb{R}_{>0},d_A)$ and $(\mathbb{R},d_B)$.
