Prove this limit $\lim \limits_{x\to\infty}f(x)=0$ I have this problem in real analysis. I think it needs integral factor or knowledge of ODE to prove, but not sure how to it. Here is the question:

Let $f$ be a real valued continuous function on $[0,\infty]$ such that
  $$
\lim \limits_{x\to\infty}\left(f(x)+\int_{0}^{x}f(t)dt\right)
$$
  exists. Prove that
  $$
\lim \limits_{x\to\infty}f(x)=0
$$

 A: We are given $f(x) + \int_0^x f \to L.$ To show $f(x)\to 0,$ we need to show $ \int_0^x f \to L.$ But note
$$\tag 1  \int_0^x f =\frac{e^x\int_0^x f}{e^x}.$$
Since the denominator on the right $\to \infty,$ we can contemplate using L'Hopital. Let's try it: The quotient of derivatives is
$$\tag 2 \frac{e^x(f(x)+\int_0^x f)}{e^x} = f(x)+\int_0^x f.$$
The right hand side of $(2)\to L$ by hypothesis. Therefore, by L'Hopital, $(1)\to L,$ and we are done.
A: Let $g(x)=e^x\int_{0}^{x}f(t)dt$, then 
$$
g'(x)=e^x\left(f(x)+\int_{0}^{x}f(t)dt\right)
$$
So there is
$$
\lim \limits_{x\to\infty}g'(x)e^{-x}=\lim \limits_{x\to\infty}\left(f(x)+\int_{0}^{x}f(t)dt\right)=A
$$
So for any $\epsilon>0$, there is $M>0$, such that for all $x>M$
$$
A-\epsilon<g'(x)e^{-x}<A+\epsilon \hspace{5 mm} \text{or} \hspace{5 mm} (A-\epsilon)e^{x}<g'(x)<(A+\epsilon)e^{x}
$$
By integrating on both side from $M$ to $x$, there is
$$
(A-\epsilon)\left(e^{x}-e^M\right)<g(x)-g(M)<(A+\epsilon)\left(e^{x}-e^M\right)
$$
So we have
$$
\left((A-\epsilon)\left(e^{x}-e^M\right)+g(M)\right)e^{-x}<\int_{0}^{x}f(t)dt<\left((A+\epsilon)\left(e^{x}-e^M\right)+g(M)\right)e^{-x}
$$
And 
$$
\varlimsup\limits_{x\to\infty}\int_{0}^{x}f(t)dt\leqslant \varlimsup\limits_{x\to\infty}\left((A+\epsilon)\left(e^{x}-e^M\right)+g(M)\right)e^{-x}=A+\epsilon
$$
$$
\varliminf\limits_{x\to\infty}\int_{0}^{x}f(t)dt\geqslant \varliminf\limits_{x\to\infty}\left((A-\epsilon)\left(e^{x}-e^M\right)+g(M)\right)e^{-x}=A-\epsilon
$$
So we have 
$$
0\leqslant \varlimsup\limits_{x\to\infty}\int_{0}^{x}f(t)dt-\varliminf\limits_{x\to\infty}\int_{0}^{x}f(t)dt\leqslant 2\epsilon
$$
Since ϵ is arbitrary, we have
$$
\varlimsup\limits_{x\to\infty}\int_{0}^{x}f(t)dt=\varliminf\limits_{x\to\infty}\int_{0}^{x}f(t)dt=A \hspace{5 mm} 
$$
Or
$$
\lim\limits_{x\to\infty}\int_{0}^{x}f(t)dt=\lim \limits_{x\to\infty}\left(f(x)+\int_{0}^{x}f(t)dt\right)
$$
So finally we have
$$
\lim \limits_{x\to\infty}f(x)=\lim \limits_{x\to\infty}\left(\left(f(x)+\int_{0}^{x}f(t)dt\right)-\int_{0}^{x}f(t)dt\right)=0
$$
