Let $f$ be a continuous function on $[0,\infty)$ such that $\lim_{x\to \infty}(f(x)+\int_0^x f(t)dt)$ exists. Find $\lim_{x\to \infty}f(x)$. Let $f$ be a continuous function on $[0,\infty)$ such that $\lim_{x\to \infty}(f(x)+\int_0^x f(t)dt)$ exists. Find $\lim_{x\to \infty}f(x)$.
Useful hints will work. Please give some nice hint to solve  the problem.
 A: One of the easiest approaches is the use of L'Hospital's Rule here. Let $F(x) =\int_{0}^{x}f(t)\,dt$ so that $F'(x) =f(x) $ via Fundamental Theorem of Calculus. We are thus given that $F(x) +F'(x) \to L$ as $x\to\infty$.
We have
\begin{align}
\lim_{x\to\infty} F(x) &=\lim_{x\to\infty} \frac{e^xF(x)} {e^x}\notag\\
&=\lim_{x\to \infty} \dfrac{e^x\{F(x) +F'(x) \}} {e^x}\text{ (via L'Hospital's Rule)} \notag\\
&=\lim_{x\to\infty} F(x) +F'(x)\notag\\
&=L\notag
\end{align}
and hence $$\lim_{x\to \infty} f(x) =\lim_{x\to\infty} F'(x) =\lim_{x\to\infty} \{F(x) + F'(x) \}-F(x) =L-L=0$$ Based on feedback received in comments, it appears that some more explanation of the steps is needed.
Note that the given hypotheses imply that limit of $$f(x) +F(x) =F'(x) +F(x) =\frac{e^x(F(x) +F'(x)) } {e^ x}=\frac{(e^xF( x)) '} {(e^x)'} $$ exists and equals $L$ as $x\to \infty $. By L'Hospital's Rule it follows that the limit of $\dfrac{e^xF(x)} {e^x} $ (which is same as $F(x) $) also exists and equals $L$. 
A: Let $F(x) = \int_0^x f(t)dt$. Then, we have $\lim_{x\to \infty} e^{-x}[e^x F(x)]' = c$ exists. From this, given $\epsilon>0$, there is $N>0$ such that for all $x>N$,
$$
(c-\epsilon)e^x \leq[e^x F(x)]' \leq (c+\epsilon)e^x.
$$
Now, integrating from $N$ to $t$ gives:
$$(c-\epsilon)(e^t-e^N)\leq e^tF(t) -e^NF(N) \leq (c+\epsilon)(e^t-e^N),\quad\forall t>N
$$and hence
$$
(c-\epsilon)(1-e^{N-t})\leq F(t) -e^{N-t}F(N) \leq (c+\epsilon)(1-e^{N-t}), \quad \forall t>N.
$$Taking $t\to\infty$ yields
$$c-\epsilon \leq \liminf_{t\to \infty}F(t)\leq \limsup_{t\to \infty}F(t)\leq c+\epsilon.
$$Since $\epsilon>0$ was arbitrary, we have $\lim_{t\to\infty} F(t) =c$ and thus $\lim_{t\to\infty} f(t) = 0$.
