Prove $f(x) = \sqrt{x}\ln(x)$ is uniformly continuous for $x = [1, \infty)$

Question

Original question is to show this is true for all $x > 0$ with the hint to split cases on $x \in (0,1]$ and $x \in [1, \infty)$. I can show this is true for $x \in (0,1]$ by extending the interval to include $0$ (because $f(0) = 0 = \lim_{x\rightarrow 0^+}f(x)$) and use Heine-Cantor Theorem. But how do I prove it for $x \in [1,\infty)$ with the $\epsilon - \delta$ definition of uniform continuity?

Here is my attempt: let $x,y \in [1,\infty)$, then for all $\epsilon > 0$, we want to find $\delta$ such that $\vert x-y \vert < \delta$ implies: $$ \begin{align*} \vert \sqrt{x}\ln(x) - \sqrt{y}\ln(y)\vert &= \vert \sqrt{x}\ln(x) - \sqrt{x}\ln(y) + \sqrt{x}\ln(y) - \sqrt{y}\ln(y)\vert \\ &\leq \sqrt{x}\,\bigg\vert\ln\left(\frac{x}{y}\right)\bigg\vert + \ln(y)(\sqrt{x} - \sqrt{y}) \\ & < \epsilon \end{align*} $$ Take cases on $x<y$ and $x\geq y$ and we can use the property that for any $x' > 0$, $1+x' < e^{x'}$ to further simplify the $\bigg\vert\ln\left(\frac{x}{y}\right)\bigg\vert$ term. But I'm not sure how to find the $\delta$ that is independent from $x,y$ from this. Any hints? Or maybe I'm on the wrong track?

Answer 1

$$f'(x)=\frac{\ln(x)}{2\sqrt{x}}+\frac{1}{\sqrt{x}}$$ When $x\to\infty$, then $f'(x)\to 0$. In particular, $f'$ is bounded on $[1,\infty)$ (since it is continuous), thus $f$ is uniformly continuous (in fact Lipschitz) by the Mean Value Theorem.

Answer 2

Since $(\forall x\in[1,\infty):f''(x)=-\frac{\log (x)}{4 x^{3/2}}<0$, $f'$ is decreasing and it attains its maximum at $1$ (and converges to $0$). So, $(\forall x\in[1,\infty):|f'(x)|\leqslant1$. So, for any $\varepsilon>0$; take $\delta=\varepsilon$ and you're done.