Can you differentiate?

My proof is a proof by contradiction:

assume that $f'(x) \underset{x \rightarrow +\infty}{\nsim} \dfrac{1}{2\sqrt{x}}$

ie $\exists \varepsilon \in ]0,1[, \forall x_0>0, \exists x_1 \geq x_0, (1-\varepsilon)\dfrac{1}{2\sqrt{x_1}}>f'(x_1) \quad \text{or} \quad f'(x_1)>(1+\varepsilon)\dfrac{1}{2\sqrt{x_1}} \quad (*)$

I set $\varepsilon$ satisfying $(*)$ .

Let $x_0 >0, \forall x \geq x_0, (1-g(\varepsilon))\sqrt{x}<f(x)<(1+g(\varepsilon))\sqrt{x} \quad (**)$ where $g$ is a function which goes to $0$ at $0$ and strictly positive (it is the hypothesis concerning $f$ ). We are going to find $g$ to contradict this statement using $(*)$ .

Let $x_1 \geq (1+\varepsilon)^2 x_0$ satisfying $(*)$ .

First case: $(1-\varepsilon)\dfrac{1}{2\sqrt{x_1}}>f'(x_1)$

let $(1-\varepsilon)\dfrac{1}{2\sqrt{x_1}} = \dfrac{1}{2\sqrt{x_2}} \Rightarrow x_2 = \dfrac{x_1}{(1-\varepsilon)^2} > x_1 \geq x_0$

$f'$ is decreasing so :

$\begin{aligned} f(x_2)&=f(x_1)+ \displaystyle\int_{x_1}^{x_2}f'(x)dx\\ &\leq f(x_1)+(x_2-x_1)\dfrac{(1-\varepsilon)}{2\sqrt{x_1}}\\ &\leq (1+g(\varepsilon))\sqrt{x_{1}}+\dfrac{\sqrt{x_{1}}}{2}(\frac{1}{1-\varepsilon}-1+\varepsilon)\\ &\leq \sqrt{x_{1}}(g(\varepsilon)+\dfrac{1+\varepsilon}{2}+\dfrac{1}{2-2\varepsilon})=(1-\varepsilon)\sqrt{x_2}(g(\varepsilon)+\dfrac{1+\varepsilon}{2}+\dfrac{1}{2-2\varepsilon}) \end{aligned}$

To contradict $(**)$ we only need that $(1-\varepsilon)(g(\varepsilon)+\dfrac{1+\varepsilon}{2}+\dfrac{1}{2-2\varepsilon}) < 1-g(\varepsilon) \ \text{ie} \ (2-\varepsilon)g(\varepsilon) < \dfrac{1}{2} - \dfrac{1-\varepsilon^2}{2} = \dfrac{\varepsilon^2}{2}$

Finally, the first condition is $g(\varepsilon)<\dfrac{\varepsilon^2}{2(2-\varepsilon)}$

Second case: $(1+\varepsilon)\dfrac{1}{2\sqrt{x_1}}<f'(x_1)$

let $(1+\varepsilon)\dfrac{1}{2\sqrt{x_1}} = \dfrac{1}{2\sqrt{x_3}} \Rightarrow x_3 = \dfrac{x_1}{(1+\varepsilon)^2} \geq x_0$

$f'$ is decreasing so :

$\begin{aligned} f(x_1)&=f(x_3)+ \displaystyle\int_{x_3}^{x_1}f'(x)dx\\ &\geq f(x_3)+(x_1-x_3)\dfrac{1+\varepsilon}{2\sqrt{x_1}}\\ &\geq (1-g(\varepsilon))\sqrt{x_3}+\dfrac{\sqrt{x_1}}{2}(1+\varepsilon - \dfrac{1}{1+\varepsilon})\\ &\geq (1-g(\varepsilon))\dfrac{\sqrt{x_1}}{1+\varepsilon} + \dfrac{\sqrt{x_1}}{2}(1+\varepsilon - \dfrac{1}{1+\varepsilon})\\ &\geq \sqrt{x_1}(\dfrac{1-g(\varepsilon)}{1+\varepsilon} + \dfrac{1+\varepsilon}{2} - \dfrac{1}{2(1+\varepsilon)}) \end{aligned}$

To contradict $(**)$ we only need that $\dfrac{1-g(\varepsilon)}{1+\varepsilon} + \dfrac{1+\varepsilon}{2} - \dfrac{1}{2(1+\varepsilon)} > 1+g(\varepsilon)$

ie $1 + \dfrac{(1+\varepsilon)^2}{2} - \dfrac{1}{2} > 1+\varepsilon + g(\varepsilon)(2+\varepsilon)$

ie $\varepsilon + \dfrac{ \varepsilon^2}{2} > \varepsilon + g(\varepsilon)(2+\varepsilon)$

ie $g(\varepsilon) < \dfrac{ \varepsilon^2}{2(2+\varepsilon)} \$ , it is the second condition

We can take $g(\varepsilon)=\dfrac{\varepsilon^2}{6}$ which ends the proof by contradiction.

BONUS: with the correct hypothesis ( $f$ concave or convexe), you can generalize this to $f: x \mapsto x^\alpha$ for $\alpha \in \mathbb{R}$ .