Uniform continuity of the function $x\log(\frac{1}{x})$ on$ (0,\infty)$.

How to check uniform continuity of the function $x\log(\frac{1}{x})$ on $(0,\infty)?$ I tried with many results . It’s not continuous periodical function. It’s derivative is not bounded. Limit doesn’t exist finitely at end points . Unable to decide whether it’s uniform continuous or not . Please help . Thanks .

$xlog(1/x)=-xlogx$. Also its derivative is $-logx -1$ which tends to $-\infty$ so your function is not uniformly continuous. (By Lagrange if you want)

This function is not uniformly continuous. Define $x=y+\delta$ where $x,y,\delta>0$ and $x$ and $y$ are sufficiently large. Therefore $${|(y+\delta)\log(y+\delta)-y\log y|<\epsilon\implies \\|y\log {1+{\delta\over y}}+\delta\log (y+\delta)|=y\log ({1+{\delta\over y}})+\delta\log (y+\delta)<\epsilon\implies\\ \delta\log(y+\delta)<\epsilon\implies\\\delta<{\epsilon\over \log y}}$$which means that given $\epsilon>0$, $\delta$ can be arbitrarily small.