Prove that $\ln(x)$ is uniformly continuous on the set $[1,\infty)$ Hello I was wondering if someone could take a look at my proof for this and see if it is sufficient.
Let $\epsilon > 0$. Choose $\delta = \epsilon$ such that $\forall x,y \in [1,\infty)$ with $|x-y| < \delta$, we have $\vert \ln x- \ln y \vert$. By the bounds of natural logarithms we know $\ln x \leq x-1~\forall x$. Thus $
\vert \ln x - \ln y \vert \leq \vert (x-1) - (y-1) \vert = \vert x-1-y+1 \vert= \vert x-y \vert \lt \delta = \epsilon$.
Thank you.
 A: The inequality $|\ln(x) - \ln(y)| \leq |(x - 1) - (y - 1)|$ does not follow from your inequality $\forall x . \ln(x) \leq x - 1$.
For consider $g(x) = x - 1 - x^2$. Then for all $x,g(x) \leq x - 1$. But we we see that $|g(3) - g(2)| = |-7 - (-3)| = 4 > 1 = |(3 - 1) - (2 - 1)|$.
But your choice of $\delta = \epsilon$ does indeed work. Hint: go back to the definition of $\ln(x) := \int\limits_1^x \frac{1}{t} dt$, and show that $|\ln(x) - \ln(y)| \leq |x - y|$ that way.
A: Let $f:X\to\mathbb{R}$ be a real-valued function. We say that $f$ is uniformly continuous iff it maps equivalent sequences onto equivalent sequences. Having said, let us consider two equivalent sequences $(x_{n})$ and $(y_{n})$, both of which lies inside $X = [1,+\infty)$, where $f(x) = \ln(x)$. Then we have
\begin{align*}
\lim_{n\to\infty}[\ln(x_{n}) - \ln(y_{n})] & = \lim_{n\to\infty}\ln\left(\frac{x_{n}}{y_{n}}\right)\\\\
& = \lim_{n\to\infty}\ln\left(\frac{x_{n} - y_{n}}{y_{n}} + \frac{y_{n}}{y_{n}}\right)\\\\
& = \lim_{n\to\infty}\ln\left(\frac{x_{n} - y_{n}}{y_{n}} + 1\right)\\\\
& = \ln\lim_{n\to\infty}\left(\frac{x_{n} - y_{n}}{y_{n}} + 1\right)\\\\
& = \ln(1) =0
\end{align*}
Then we conclude that $\ln:[1,+\infty)\to[0,+\infty)$ is uniformly continuous.
