Context: Here $\tau = \langle X, T \rangle $ and $v = \langle X, N \rangle$. I understood everything in the proof except for the blue part. Why is it that $\displaystyle{\lim_{x \to \infty} y'(x)} > 0$? Where do the final inequalities come from?
This is the thesis where I got the text. Page 25.
Update: I see why, if $\displaystyle \lim _{x \to \infty} \frac{y(x)}{x} = A$ for some constant $A$, and $\dfrac{y(x)}{x}$ is decreasing then $y(x) - Ax > 0 $, but I still do not see why $A$ has to be a constant (why cannot that limit blow up to infinity?) or why it is the case that$$ y(x) - Ax \leq x \int_x^\infty \frac{w(t)}{t^2} \,\mathrm{d}t \leq \alpha e^{-\frac{x^2}{2}}?$$
Edit: Bounty added.
$\def\d{\mathrm{d}}$First, $w(x) > 0 \Rightarrow \dfrac{y(x)}{x} > y'(x)$ for all $x > 0$. For $x > 1$, because that $\dfrac{y(x)}{x}$ is decreasing, then $y'(x) < \dfrac{y(x)}{x} \leqslant y(1)$. Also, $y'(x)$ is increasing, so $A = \lim\limits_{x → +∞} y'(x)$ exists and is finite. Since $\lim\limits_{x → +∞} w(x) = 0$, then$$ \lim_{x → +∞} \frac{y(x)}{x} = \lim_{x → +∞} y'(x) + \lim_{x → +∞} \frac{w(x)}{x} = A + 0 = A. $$
Next, because $\dfrac{y(x)}{x}$ is strictly decreasing and $\lim\limits_{x → +∞} \dfrac{y(x)}{x} = A$, then $\dfrac{y(x)}{x} > A \Rightarrow y(x) - Ax > 0$ for $x > 0$. Note that $\left( \dfrac{y(x)}{x} \right)' = -\dfrac{w(x)}{x^2}$, thus$$ \int_x^{+∞} \frac{w(t)}{t^2} \,\d t = -\left. \frac{y(t)}{t} \right|_x^{+∞} = \frac{y(x)}{x} - A. $$ Also because $y'(x)$ is increasing and $\lim\limits_{x → +∞} y'(x) = A$, then $y'(x) \leqslant A$ for $x > 0$, Therefore,$$ x \int_x^{+∞} \frac{w(t)}{t^2} \,\d t = y(x) - Ax \leqslant y(x) - x y'(x) = w(x) \leqslant α\exp\left( -\frac{x^2}{2} \right). $$