Let $h$ -- positive, twice continuously differentiable function on the $x\ge0$, such that $h'(x)>0$ and $h''(x)=o((h'(x))^2),$ $x\to\infty$. Show that $\displaystyle\int\limits_0^\infty e^{-h(t)}dt$ converges.
My attempt $$\int\limits_{x_0}^A e^{-h(t)}dt=-\int\limits_{x_0}^A \frac{de^{-h(t)}}{h'(t)}=\frac{e^{-h(x_0)}}{h'(x_0)}-\frac{e^{-h(A)}}{h'(A)}-\int\limits_{x_0}^A e^{-h(t)}\frac{h''(t)}{(h'(t))^2}dt=$$$$=\frac{e^{-h(x_0)}-e^{-h(A)}}{h'(x_0)}-e^{-h(A)}\left(\frac{1}{h'(A)}-\frac{1}{h'(x_0)}\right)-\int\limits_{x_0}^A e^{-h(t)}\frac{h''(t)}{(h'(t))^2}dt=$$$$=\frac{e^{-h(x_0)}-e^{-h(A)}}{h'(x_0)}+e^{-h(A)}\frac{h''(\xi)}{(h'(\xi))^2}(A-x_0)-\frac{h''(\eta)}{(h'(\eta))^2}\int\limits_{x_0}^A e^{-h(t)}dt.$$ Using that $\left|\dfrac{h''(x)}{(h'(x))^2}\right|\le\dfrac{1}{2}$ for $x\ge x_0$, i get $$\int\limits_{x_0}^A e^{-h(t)}dt\le\frac{e^{-h(x_0)}-e^{-h(A)}}{h'(x_0)}+\frac{1}{2}e^{-h(A)}(A-x_0)+\frac{1}{2}\int\limits_{x_0}^A e^{-h(t)}dt,$$$$\int\limits_{x_0}^A e^{-h(t)}dt\le2\frac{e^{-h(x_0)}-e^{-h(A)}}{h'(x_0)}+e^{-h(A)}(A-x_0).$$ The first fraction has a finite limit, because $h(A)$ is increasing. In the second term $$e^{-h(A)}=e^{-h(x_0)}e^{-h'(x_0)(A-x_0)}e^{-h''(\zeta)(A-x_0)^2/2}\le e^{-h(x_0)}e^{-h'(x_0)(A-x_0)}e^{-(h'(\zeta))^2(A-x_0)^2/4}\le e^{-h(x_0)}e^{-h'(x_0)(A-x_0)},$$ so the second term tends to $0$.
Is my reasoning correct? I'm confused by the fact that I have'nt used the property $\dfrac{h''(x)}{(h'(x))^2}\to0$. Maybe there is a simpler solution using this property?
