Show that $\lim_{x \to +\infty}\left(f(x)+f'(x)\right)=0 \Rightarrow \lim_{x \to +\infty} f(x)=0$ How to show that $\lim_{x \to +\infty}(f(x)+f'(x))=0 $ implies $\lim_{x \to +\infty} f(x)=0$?
 A: If $\lim\limits_{x\to+\infty} f(x) + f'(x) = 0$, then for any $\epsilon > 0$, there exists
$y > 0$ such that for all $x > y$,
$$\begin{align}
& f(x) + f'(x) \le \epsilon\\
\implies & \frac{d}{dx}\big[f(x) e^x\big] = (f(x)+f'(x)) e^x \le \epsilon e^x\\
\implies & f(x)e^x - f(y)e^y = \int_y^x \frac{d}{dt}\big[f(t)e^t\big]dt \le \epsilon \int_y^x e^t dt = \epsilon( e^x - e^y)\\
\implies & f(x) \le \epsilon + ( f(y) - \epsilon ) e^{y-x}\\
\end{align}
$$
This leads to 
$$\limsup_{x\to+\infty} f(x) \le \epsilon + |f(y)-\epsilon|\lim_{x\to+\infty} e^{y-x} = \epsilon \tag{*1a}$$
Apply exactly the same argument to $-f(x)$, we get
$$\limsup_{x\to+\infty} ( -f(x) ) \le \epsilon\quad\implies\quad
\liminf_{x\to+\infty} f(x) \ge -\epsilon \tag{*1b}$$
Since $\epsilon$ is arbitrary, $(*1a)$ and $(*1b)$ together implies
$$0 = 
\underbrace{\sup_{\epsilon>0} (-\epsilon ) \le \liminf_{x\to+\infty} f(x)}_{(*1b)}
\;\le\;
\underbrace{\limsup_{x\to+\infty}f(x) \le \inf_{\epsilon > 0}\epsilon}_{(*1a)}
= 0
\quad\implies\quad
\lim_{x\to+\infty} f(x) = 0\\
$$
A: Hint: If, $L = \displaystyle \lim\limits_{x \to \infty} f(x) = \lim\limits_{x \to \infty}\dfrac{e^xf(x)}{e^x}$ and since $\lim\limits_{x \to \infty} e^x = \infty$, L-Hopital suggests,
$\displaystyle L = \lim\limits_{x \to \infty} \dfrac{(e^xf(x))'}{(e^x)'} = \lim\limits_{x \to \infty} f(x) + f'(x) = 0$
we still need to establish the existence of the limits !
Another way is using the Cauchy MVT.
For a $\epsilon > 0$, choose an an $\alpha > 0$, such that $\displaystyle |f(x) + f'(x)| \le \epsilon$, for $x \ge \alpha$.
Now Cauchy's Mean Value Theorem implies $\exists \beta \in (\alpha,x)$, such that
$$\displaystyle \dfrac{e^xf(x) - e^{\alpha}f({\alpha})}{e^x - e^{\alpha}} = f(\beta)+f'(\beta)$$
It follows that $|f(x) - f(a)e^{\alpha-x}| \le \epsilon|1-e^{\alpha-x}| \implies |f(x)| \le |f(\alpha)|e^{\alpha-x}+\epsilon|1-e^{\alpha-x}|$
Hence, $|f(x)| \le 2\epsilon$, for large enough $x$.
A: This was already posted on the site, but here we go: consider the function $g$ defined by $$g(x)=\mathrm e^xf(x),$$ then $$g'(x)=\mathrm e^x(f'(x)+f(x)),$$ hence, by hypothesis, for every positive $\varepsilon$, for every $x$ large enough, say, every $x\geqslant x_\varepsilon$, $$|g(x)|\leqslant\varepsilon\mathrm e^x.$$ Integrating this on the interval $(x_\varepsilon,x)$, one gets $$|g(x)|\leqslant|g(x_\varepsilon)|+\varepsilon(\mathrm e^x-\mathrm e^{x_\varepsilon})\leqslant|g(x_\varepsilon)|+\varepsilon\mathrm e^x,$$ for every $x\geqslant x_\varepsilon$, that is, $$|f(x)|\leqslant\varepsilon+|g(x_\varepsilon)|\mathrm e^{-x}.$$
When $x\to+\infty$, this yields $$\limsup_{x\to+\infty}|f(x)|\leqslant\varepsilon.$$ This upper bound holds for every positive $\varepsilon$ hence $$\lim_{x\to+\infty}|f(x)|=0.$$
A: Notice that we can, from the limit, for any $\varepsilon>0$ we can write that, for all large enough $x$ tht
$$-\varepsilon< f(x)+f'(x)<\varepsilon$$
which can be written as
$$-\varepsilon-f(x) < f'(x) < \varepsilon-f(x)$$
which, as a simple bound, implies that if $f(x)>2\varepsilon$ then $f'(x)<-\varepsilon$. This means that $f$, for large enough $x$, is definitely eventually going to less than $2\varepsilon$. You can expand this into a full proof that the limit $\lim_{x\rightarrow\infty}f(x)$ exists and is $0$ easily.
A: Remarks: 1. I was supposed to answer this problem (2020/04/20) but was closed: Prove if $\lim_{x\to +\infty}(f(x)+f'(x))=0$ then $\lim_{x\to +\infty} f(x)=0$


*The usual L'Hopital's rule for $\frac{\infty}{\infty}$
does not work directly for the problem since one has to prove that
$\lim_{x\to \infty} \mathrm{e}^xf(x) = \infty$.


*It seems the following version of L'Hopital's rule is not used directly in the answers here. So I wrote down this answer.


*The usual L'Hopital's rule deals with $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
As a discrete version of L'Hopital's rule, the Stolz-Cesaro theorem deals with $\frac{0}{0}$ or $\frac{\cdot}{\infty}$.
Why doesn't the usual L'Hopital'rule deal with $\frac{\cdot}{\infty}$ like the Stolz-Cesaro theorem?
$\phantom{2}$
Proof
We will use the following version of L'Hopital's rule (page 3 in [1]):
Suppose $f(x)$ and $g(x)$ are differentiable on an open interval $I$ except possibly at a point $A$
contained in $I$, $g'(x)\ne 0$ for all $x$ in $I$, $\lim_{x\to A} g(x) = \infty$,
and $\lim_{x\to A} \frac{f'(x)}{g'(x)} = L$. Then:
$$\lim_{x\to A} \frac{f(x)}{g(x)} = L.$$
(Remark: Nothing is assumed about $\lim_{x\to A} f(x)$, not even its existence.)
$\phantom{2}$
By using L'Hopital's rule above,
since $\lim_{x\to \infty} \frac{(\mathrm{e}^x f(x))'}{(\mathrm{e}^x)'}
= \lim_{x\to \infty} f(x) + f'(x) = 0$
and $\lim_{x\to \infty} \mathrm{e}^x = \infty$,
we have $\lim_{x\to \infty} \frac{\mathrm{e}^x f(x)}{\mathrm{e}^x} = \lim_{x\to \infty} f(x) = 0$.
We are done.
Reference
[1] Gabriel Nagy, https://www.math.ksu.edu/~nagy/snippets/stolz-cesaro.pdf
