Uniform continuity of $f(x) = \frac{\log{x}}{x}$ in $\left [1,+\infty \right )$.

Question

It is asked to discuss the uniform continuity of $f(x):=\frac{\log{x}}{x}$ in the interval $\left [1,+\infty \right ) \subset \mathbb{R}$, with the usual metric.

The first thing I've done is supposing that for every $x,y\in\mathbb{R}$, it is $\left |x-y\right | < \delta :=\varepsilon$, where $\varepsilon >0$. Then, I have tried to see that $\left |\frac{\log{x}}{x} - \frac{\log{y}}{y}\right |$ is bounded by $\varepsilon$ or some multiple of it. However, I have only reached the following chain of equalities and inequalities $$ \left |\frac{\log{x}}{x} - \frac{\log{y}}{y}\right | = \left | \frac{y\log{x}-x\log{y}}{xy} \right | \underset{x,y \geq 1}{=} \frac{\left |y\log{x}-x\log{y}\right |}{xy} = \frac{\left |y\log{x}-xy+xy-x\log{y}\right |}{xy} \leq \frac{ |y| |\log{x}-x |+ |x||\log{y}-y|}{xy} $$

which I don't know how to bound.

Any kind of help will be appreciated.

Answer 1

Since $f'(x)=\frac{1-\log x}{x^2}\in(-1,1]$ for every $x\geqslant1$, you have, by the mean value theorem, that $\bigl\lvert f(y)-f(x)\bigr\rvert\leqslant\lvert y-x\rvert$ for every $x$ and every $y$ in $[1,\infty)$. So…

Answer 2

$\lim_{x \to \infty} f(x)=0$ by L'Hopital's Rule. Any continuous function on $[1,\infty)$ which has a finite limit at $\infty$ is uniformly continuous.

Alternatively, you can show that $f'(x)$ is bounded. By MVT this implies that $f$ is uniformly continuous.

Answer 3

Here, on the given interval derivative of f is bounded. Hence, it is uniformly continuous.

Also, on the given interval f is defined and continuous everywhere, and also the limit as x goes to infinity also exists, hence it is uniformly continuous.