$L=\lim_{x\to\infty}(f(x)+f'(x))$ exists . Which of the following statements is\are correct? Let $f$ be a continously differentiable function on $\mathbb R$.  Suppose that 
$$L=\lim_{x\to\infty}(f(x)+f'(x))$$ exists.  If $0<L<\infty$, then which of the following statements is\are correct?


    
*
    
*If $\lim_{x\to\infty} f'(x)$ exists, then it is $0$.
    
*If $\lim_{x\to\infty} f(x)$ exists, then it is $L$.
    
*If $\lim_{x\to\infty} f'(x)$ exists, then $\lim_{x\to\infty}f(x)=0$.
    
*If $\lim_{x\to\infty} f(x)$ exists, then $\lim_{x\to\infty}f'(x)=0$.


My Guess
I could not conclude the answer and prove that properly. But, I guess that it must be 1 and 2. help me.
 A: Hint:
If $\lim f'(x) = M$, then $\lim f(x) = L-M$
Use the MVT: $f(x+1) - f(x) = f'(\xi)$ with $x < \xi < x+1$.  
A: *

*If $\lim\limits_{x\to\infty}f'(x)$ exists, then $\lim\limits_{x\to\infty}f(x)=L-\lim\limits_{x\to\infty}f'(x)$ also exists. Then
$$
\begin{align}
\lim\limits_{x\to\infty}f'(x)
&=\lim\limits_{x\to\infty}(f(x+1)-f(x))\\
&=\lim\limits_{x\to\infty}f(x+1)-\lim\limits_{x\to\infty}f(x)\\
&=0\tag{1}
\end{align}
$$

*If $\lim\limits_{x\to\infty}f(x)$ exists, then $(1)$ implies that $\lim\limits_{x\to\infty}f'(x)=0$ and therefore,
$$
\begin{align}
\lim\limits_{x\to\infty}f(x)
&=L-\lim\limits_{x\to\infty}f'(x)\\
&=L\tag{2}
\end{align}
$$

*If $\lim\limits_{x\to\infty}f'(x)$ exists, then $(1)$ and $(2)$ say that $\lim\limits_{x\to\infty}f'(x)=0$ and $\lim\limits_{x\to\infty}f(x)=L$.

*If $\lim\limits_{x\to\infty}f(x)$ exists, then $(1)$ implies that $\lim\limits_{x\to\infty}f'(x)=0$.
A: You are correct about 1 and 2.
Note also that 2 implies 4, since
$$
\lim_{x \to \infty} f'(x) = L - \lim_{x \to \infty} f(x)
$$
(assuming the latter limit exists).

Note that for arbitrary functions $g,h$: if $\lim_{x \to \infty} g(x)$ and $\lim_{x \to \infty} h(x)$ both exist, then
$$
\lim_{x \to \infty} [g(x) \pm h(x)] = 
\lim_{x \to \infty} g(x) \pm \lim_{x \to \infty} h(x)
$$
A: Hint: Check this: $$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty}\frac{e^xf(x)}{e^x} \color{red}{=} \lim_{x \to +\infty}\frac{e^x(f(x)+f'(x))}{e^x} = \lim_{x \to +\infty}f(x)+f'(x) = L,$$ by $\color{red}{\text{L'Hospital's rule}}$.
