Prove that $f'(x) \to 0$ as $x \to \infty$ when $f'+f''$ is bounded abve. 
I am given $f \in C^2([0,\infty))$ and $\lim_{x \to \infty}f(x) = L
 \in \mathbb{R}$. I am also given there is a real number $M$ such that
  $f'(x) + f''(x) < M$ for all $x \in [0,\infty)$. Prove that $\lim_{x
 \to \infty}f'(x) = 0$.

My thoughts 
This is similar to a frequently asked question: If $\lim_{x \to \infty}f(x) = L$ then if $\lim_{x \to \infty} f'(x)= L'$ exists it is necessary that $L' = 0$. Possible approaches are:
(1) Originally I thought to show upper bound on $f' + f''$ along with $f(x) \to L$ guarantees that the limit of $f'$ exists. but I just realized the function $f(x) = \sin(x^2)/x$ is a counterexample.
(2) Somehow use the Taylor expansion $f(y) = f(x) + f'(x)(y-x) + \frac{1}{2} f''(\xi)(x-y)^2$ ($\xi \in (x,y)$). Here I am having difficulty tying $f'$ and $f''$ together to use the bound.
 A: Rearranging the Taylor expansion with the integral remainder,
$$f(x+h) = f(x) + f'(x) h + \int_x^{x+h} (x+h-t)f''(t) \,  dt,$$
we get
$$f'(x)h = f(x+h) - f(x)+\int_x^{x+h} (x+h-t)f'(t)\,  dt - \int_x^{x+h} (x+h-t)[f'(t) + f''(t)]\,  dt$$
Using integration by parts and the mean value theorem with $\theta \in (0,1)$, the first integral is
$$\int_x^{x+h} (x+h-t)f'(t)\,  dt = f(x)h - \int_x^{x+h}f(t) \, dt = f(x)h-f(x+\theta h)h,$$
and the second integral is bounded as
$$ \int_x^{x+h} (x+h-t)[f'(t) + f''(t)]\,  dt \leqslant M\int_x^{x+h} (x+h-t)\,  dt = \frac{Mh^2}{2}$$
Altogether we have
$$f'(x)h \geqslant f(x+h) - f(x) +f(x)h - f(x+\theta h) h - \frac{Mh^2}{2}$$
Dividing both sides by $h \in (0,1)$, we get $\liminf_{x \to \infty}f'(x) \geqslant -\frac{Mh}{2}$ since the limits of all terms involving $f$ cancel out. This implies $\liminf_{x \to \infty}f'(x) \geqslant 0$ since $h$ can be arbitrarily close to $0$.
Similarly, dividing by $h \in (-1,0)$ we can show that $\limsup_{x \to \infty}f'(x) \leqslant 0$, and therefore, $\lim_{x \to \infty} f'(x) = 0$. 
