Asymptotic behaviour of $f$ and $f'+f$ 
Let $f~:\mathbb{R}\longrightarrow \mathbb{R}$ a $C^1$ map. How do you prove that if $f'+f$ vanish when $t\to +\infty$ so do $f$?

If $\lim\limits_{t\to +\infty} f(t)=l$ with $l\neq 0$ :
I first look at the case $l$ finite : we have $\lim\limits_{t\to +\infty} f'(t)=-l$ so for every $\varepsilon >0$ there exist $C>0$ such for all $t>C$, $|f(t)-l|$ and $|f'(t)+l|$ are both $\leq \varepsilon$. But I can't find a contradiction.
We know that $f'$ has a limit in $+\infty$ does not implies that there exist an asymptote so I don't really know what to do.
I also tried to write $g=f+f'$ and write $f$ has a solution of a differantial equation on $g$.
Any help will be appriaciated.
 A: Consider the function $h \colon t \mapsto f(t)e^t$. The assumption on $f$ says that for every $\varepsilon > 0$ there is an $x$ such that $\lvert h'(t)\rvert \leqslant \varepsilon e^{t}$ for $t \geqslant x$. Then, for $y > x$ we have
$$\lvert h(y)\rvert = \Biggl\lvert h(x) + \int_x^y h'(t)\,dt\Biggr\rvert \leqslant \lvert h(x)\rvert + \varepsilon \int_x^y e^t\,dt < \lvert h(x)\rvert + \varepsilon e^y,$$
and hence
$$\lvert f(y)\rvert = \lvert h(y)\rvert e^{-y} < \lvert h(x)\rvert e^{-y} + \varepsilon.$$
Thus, there is a $z$ such that $\lvert f(y)\rvert \leqslant 2\varepsilon$ for all $y \geqslant z$, and this shows
$$\lim_{t\to +\infty} f(t) = 0.$$
A: There is a neat trick for this specific question (and the likes of it): Consider the equality $$\lim_{x\rightarrow \infty} f(x) = \lim_{x\rightarrow \infty} f(x)  \frac{e^x}{e^x}$$
The differentiability of $f$ allows us to attempt to calculate the limit using l'Hopital's rule. We differentiate the RHS to obtain:
$$\text{RHS} \underbrace{=}_{?} \lim_{x\rightarrow \infty} \left(f(x)+f'(x)\right)  \frac{e^x}{e^x} =\lim_{x\rightarrow \infty} f(x)+f'(x) $$
with the first equality holding only if the latter limit exists, and it does.
A: Let
$f'(t) + f(t) = g(t); \tag 1$
then we have
$\lim_{t \to \infty} g(t) = 0, \tag 2$
and we see that (1) is in fact an ordinary differential equation with, for any $t_0 \in \Bbb R$, $f(t_0)$ the initial condition at $t = t_0$.  It is well-known that the unique solution to (1) is given by
$f(t) = e^{-(t - t_0)}(f(t_0) + \displaystyle \int_{t_0}^t e^{(s - t_0)}g(s) ds); \tag 3$
from (3) we have
$\vert f(t) \vert= \vert e^{-(t - t_0)}(f(t_0) + \displaystyle \int_{t_0}^t e^{(s - t_0)}g(s) ds) \vert$
$= e^{-(t - t_0)} \vert f(t_0) + \displaystyle \int_{t_0}^t e^{(s - t_0)}g(s) ds \vert \le e^{-(t - t_0)}\vert f(t_0) \vert +  e^{-(t - t_0)} \vert \int_{t_0}^t e^{(s - t_0)}g(s) ds \vert; \tag 4$
as for the rightmost integral in (4),
$\displaystyle \vert \int_{t_0}^t e^{(s - t_0)}g(s) ds \vert \le \int_{t_0}^t e^{(s - t_0)} \vert g(s) \vert ds, \tag 5$
and if $\vert g(t) \vert \le M$ for $t \ge t_0$, 
$\displaystyle \int_{t_0}^t e^{(s - t_0)} \vert g(s) \vert ds \le M\int_{t_0}^t e^{(s - t_0)} = M(e^{(t - t_0)} - 1) \tag 6$
for $t \ge t_0$ as well.  Thus
$ e^{-(t - t_0)} \vert \displaystyle \int_{t_0}^t e^{(s - t_0)}g(s) ds \vert \le Me^{-(t - t_0)} (e^{(t - t_0)} - 1) = M(1 - e^{-(t - t_0)}) \le M; \tag 7$
we combine (4) with (7) and obtain
$\vert f(t) \vert \le e^{-(t - t_0)} \vert f(t_0) \vert +  M(1 - e^{-(t - t_0)}) \le   e^{-(t - t_0)} \vert f(t_0) \vert +  M, \tag 8$
again, valid for $t \ge t_0$, provided $\vert g(t) \vert \le M$ for such $t$.
Based upon (2), given $\epsilon > 0$, we may now choose $\tau_0$ sufficiently large that
$\vert g(t) \vert \ \le \dfrac{\epsilon}{2} \; \text{for} \; t \ge \tau_0; \tag 9$
then (8) yields
$\vert f(t) \vert \le  e^{-(t - \tau_0)} \vert f(\tau_0) \vert + \dfrac{\epsilon}{2}  \; \text{for} \; t \ge \tau_0;  \tag{10}$
we may now choose $\tau_1 > \tau_0$ sufficiently large that
$ e^{-(t - \tau_0)}\vert f(\tau_0) \vert \le \dfrac{\epsilon}{2} \; \text{for} \; t \ge \tau_1, \tag{11}$
so (10) becomes
$\vert f(t) \vert \le \dfrac{\epsilon}{2} + \dfrac{\epsilon}{2} = \epsilon \; \text{for} \; t \ge \tau_1.  \tag{12}$
We have now shown that, for any $\epsilon >0$ there exists a $\tau_1> 0$ such that, if $t \ge \tau_1$, 
$\vert f(t) \vert \le \epsilon, \tag{13}$
which is precisely the assertion that
$\lim_{t \to \infty} f(t) = 0. \tag{14}$
