$\lim_{x\to+\infty}\frac{f(x)}{e^x}=l$, then $\lim_{x\to+\infty}\frac{f'(x)}{e^x}=l$? 
Let $f(x)$ be a  function,  the second derivative $f^{\prime\prime}(x)$ exists, and $\lim\limits_{x\to +\infty}\dfrac{f(x)}{e^x}=l$.  $c\gt0$ such that  for sufficiently large $x$,  $|f''(x)|<c|f'(x)|$. then we have

$$\lim_{x\to + \infty}\frac{f'(x)}{e^x}=l $$


converse of L'Hôpital's rule ? 
Thank you very much for your help
 A: I think I got it :
lets define $G(x) = f(x)/e^{x} $ we have $G'(x) = \frac {f'(x)}{e^{x}} - \frac {f(x)}{e^{x}}$


*

*for $l < \infty $ :


Suppose $  \lim_{x\to + \infty}G'(x) \neq 0 $
this means that $ \exists\ \epsilon > 0 $ for which we can find a monotone sequence ($x_{n}) $ validating : $ |G(x_{n})| > \epsilon $
The mean value theorem allows use to find a sequnce $ c_{n} \in [x_{n},x_{n}+1] $  so that :
$$ G(x_{n}+1) - G(x_{n}) = G'(c_{n})(x_{n}+1-x_{n}) = G'(c_{n}) $$
Which gives us :
$$ G(x_{n}+1) - G(x_{n}) = \frac {f'(c_{n})}{e^{c_{n}}} - \frac {f(c_{n})}{e^{c_{n}}} $$
The first par tends toward $0$, so $\frac {f'(c_{n})}{e^{c_{n}}} \sim l$.
The inequality $ |f''(x)|<c|f'(x)| $ and the mean theorem allows use to say that $ f' $ is continue.
since $c_{n}\sim x_{n}$, $G(x_{n}) \sim 0 $. Absurde.


*

*for $l = +\infty $ :


Suppose $\lim_{x\to + \infty}\frac {f'(x)}{e^{x}} \neq +\infty$.
$ \exists\ M > 0 $ for which we can find a monotone sequence ($x_{n}) $ so that : $ \frac {f'(x_{n})}{e^{x_{n}}} < M $.
Thus $ G'(x_{n}) = \frac {f'(x_{n})}{e^{x_{n}}} - \frac {f(x_{n})}{e^{x_{n}}}  \sim -\infty $. Starting some range $m $, $ n \geq m \Rightarrow  G'(x_{n}) < 0 $ (because of the continuity).
The mean theorem again,   $ c_{n} \in [x_{m},x_{n}] $  so that :
$$ G(x_{n}) - G(x_{m}) = G'(c_{n})(x_{n}-x_{m}) $$ and $$ G'(c_{n}) < 0 $$
We tend n to $+\infty$, wich gives : $G(x_{n})$ tends to $-\infty$. Absurde.


*

*For $l = -\infty$ : Take $f = -f$.


I hope it's correct.
A: Note that $\lim_{x \rightarrow +\infty} \frac{f(x+1) - f(x)}{e^{x+1} - e^x}=l$ since 
$$ \frac{f(x+1) - f(x)}{e^{x+1} - e^x}=\frac{1}{e^{x+1}} \frac{f(x+1) - f(x)}{1 - e^{-1}} $$
$$= \left[ \frac{f(x+1)}{e^{x+1}} - \frac{1}{e} \frac{f(x)}{e^x} \right] \left( \frac{1}{1-e^{-1}} \right) $$
$$\rightarrow \left[ l - \frac{l}{e} \right]\left( \frac{1}{1 - e^{-1}} \right),  \ x \rightarrow +\infty$$
$$=l$$
By MVT, there is a $x < y < x + 1$ such that $\frac{f(x+1) - f(x)}{e^{x+1} - e^x}=\frac{f'(y)}{e^y}$. 
Suppose $c \le 1$. By hypothesis, there is an $A>0$ such that $|f'(x)|>|f''(x)|$ if $x > A$. In particular, $f'$ has no zeros for $x>A$, so let's suppose $f'>0$. 
Let $h(x)=\frac{f'(x)}{e^{x}}$. Since $\left| \frac{d}{dx} \log(f'(x)) \right| = \left| \frac{f''(x)}{f'(x) } \right| < 1$, we have that $\log(f'(x)) - 1 = \log h(x)$ is Lipschitz with constant $1$, by the MVT. It follows that $h$ is decreasing and positive, so $\lim_{x \rightarrow +\infty} h(x)$ exists.
Given a positive integer $n$, take $n < y_n < n + 1$ such that $\frac{f(n+1) - f(n)}{e^{n+1} - e^n}=h(y_n)$.
Therefore, $\lim_{x \rightarrow +\infty}\frac{f'(x)}{e^x}=\lim h(y_n)=l$.
For $c>1$, apply the result for $g(x)=f(\frac{x}{c})$. (Not so sure about this piece.)
