If $\lim\limits_{x\rightarrow\infty} (f'(x)+f(x)) =L<\infty$, does $\lim\limits_{x\rightarrow\infty} f(x) $ exist? I want to prove or disprove this problem:
If there exist $\lim\limits_{x\rightarrow \infty} (f'(x)+f(x))=L&lt;\infty$ then $\lim\limits_{x\rightarrow\infty} f(x) =L$.

When I assume problem below:

If there exist $\lim\limits_{x\rightarrow\infty} (f'(x)+f(x)) =L&lt;\infty$, There exists $\lim\limits_{x\rightarrow\infty} f(x)$?

I can use mean-value theorem to show that.

So my question is:


  If $\lim\limits_{x\rightarrow\infty} (f'(x)+f(x))=L&lt;\infty$, does $\lim\limits_{x\rightarrow\infty} f(x)$ exist?
Consider the function $$g(x)=e^{x} f(x).$$ Then $$Dg(x)=e^{x}f(x)+e^{x}Df(x)=e^{x} \left( f(x)+Df(x) \right).$$
Now,
$$
\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} \frac{g(x)}{e^x} = \lim_{x \to +\infty} \frac{Dg(x)}{e^x} = \lim_{x \to +\infty} Df(x)+f(x)
$$
by De l'Hospital's theorem.

N.B. I think this exercise was solved by G. Hardy in one of his books.
When $f'$ is continuous, you can avoid l'Hôspital's rule by saying:

Let $\epsilon &gt;0$. There exists $x_0 \in \mathbb{R}$ such that $x \geq x_0$ implies $$L-\epsilon &lt; f'(x)+f(x) &lt;L+\epsilon$$

Hence $$(L-\epsilon)e^x &lt; (f'(x)+f(x))e^x &lt;(L+\epsilon)e^x$$
$$\int_{x_0}^t(L-\epsilon)e^xdx &lt; \int_{x_0}^t (f'(x)+f(x))e^xdx &lt;\int_{x_0}^t (L+\epsilon)e^xdx$$
$$(L-\epsilon)e^t-(L-\epsilon)e^{x_0} &lt; e^tf(t)-f(0)&lt;(L+\epsilon)e^t-(L+ \epsilon)e^{x_0}$$
$$(L-\epsilon) - (L-\epsilon)e^{x_0-t} &lt; f(t)-f(0)e^{-t} &lt; (L+ \epsilon)-(L+\epsilon)e^{x_0-t}$$

for $t \geq x_0$. When $t$ is large enough, we get $$L-2\epsilon &lt; f(t) &lt; L+2\epsilon$$

Therefore, $\lim\limits_{x \to + \infty} f(x)=L$.