Is it necessary that if $\lim_{x \to \infty } f(x) + f'(x) = L$ , where $L$ is finite then $\lim_{x\to\infty}f(x)= L$ and $\lim_{x\to\infty}f'(x)=0$ 
Let $f(x)$ be differentiable in $x \in (0,\infty)$ and suppose $$\lim_{x \to \infty } f(x) + f'(x)  = L$$ , where $L$ is finite quantity then find $$\lim_{x \to \infty } f(x)  \text{ and } \lim_{x \to \infty } f'(x)$$ or can nothing definitive be said about these limits.

Intuitively it seems that $\lim_{x \to \infty } f'(x) = 0 $ because if it is not , as we approach $\infty$ , its value will keep on changing as we go farther towards right. And thus $\lim_{x \to \infty } f(x) = L $. But how to prove that this happens or if it is really this that happens. 
Eg like in the function $x\sin \frac{1}{x}$, the limit approaches $0$ but it's derivative keeps oscillating. So maybe , something similar might happen at $\infty$ where the function approaches a limit but it's derivative keeps oscillating.
How do we solve it ?? Please help 
Edit: As it has been pointed out that this is a possible duplicate, I must say, that the  question  does not have a satisfactory answer and consider the follow up argument-
If we consider the function $\frac {e^x f(x) }{e^x} $ And apply L' Hopital's Rule, as @ΜάρκοςΚαραμέρης points out Even in the case if the limit of $f(x)$ exists consider for example $f(x)=e^{-x}sin(\frac{1}{x})$, how is L Hospital applicable now? You need a formal proof that $\lim_{x\to\infty}e^xf(x)$ exists
Also as @PeterForeman point out, in the accepted answer it is assumed that $f'(x) \to 0$, instead of proving it.
 A: Consider $r(x) := f'(x) + f(x)$ where we will view this as being a fixed function, and $f'(x) + f(x) = r(x)$ as giving a differential equation.  The solution to this ODE is $f(x) = e^{-x} \int e^x r(x) \, dx$.
Now, given that $r(x) \to L$ as $x \to \infty$, we have that $\int e^x r(x)\,dx = e^x L + o(e^x)$, so $f(x) = L + o(1)$.  Therefore, $\lim_{x\to \infty} f(x) = L$, and $\lim_{x\to \infty} f'(x) = \lim_{x\to \infty} (r(x) - f(x)) = L - L = 0$.
(To see how to prove the claim that $r(x) \to L$ as $x \to \infty$ implies $\int e^x r(x)\,dx = e^x L + o(e^x)$, it is fairly easy to reduce to the case $L = 0$ by using $r(x) - L$ in place of $r(x)$.  Now, for each $\epsilon > 0$, suppose that $|r(x)| < \epsilon$ whenever $x > R$; then $\int e^x r(x) \,dx = C + \int_R^x e^t r(t)\,dt$, and
$$\left|\int_R^x e^t r(t)\,dt\right| \le \int_R^x e^t |r(t)|\,dt \le \int_R^x e^t \cdot \epsilon \, dt = \epsilon(e^x - e^R).$$
From here, it should be straightforward to see that $\left|\int e^x r(x)\,dx\right| < 2\epsilon e^x$ for $x$ sufficiently large; and since this is true for any $\epsilon > 0$, we can conclude the desired result that $\int e^x r(x)\,dx = o(e^x)$.)
A: First, replacing $f(x)$ by $f(x)-L$, we can reduce to the case that $L=0$. So we're given $f(x)+f'(x)\to0$ as $x\to\infty$, and the objective is to prove that $f(x)\to0$ as $x\to\infty$. It suffices to show that, given any $\varepsilon>0$, we have $f(x)<\varepsilon$ for all sufficiently large $x$. Indeed, if we can do this, then the same argument applied to $-f$ gives $f(x)>-\varepsilon$ and therefore $|f(x)|<\varepsilon$ for all sufficiently large $x$.
So let an arbitrary $\varepsilon>0$ be given. By assumption, we can fix some number $A$ such that $f(x)+f'(x)<\varepsilon/3$ for all $x>A$. Of course, if $f(x)<\varepsilon$ for all $x>A$, then we have what we need.
So suppose that we have some $B>A$ with $f(B)\geq\varepsilon$. Since we have, by our choice of $A$, $f(B)+f'(B)<\varepsilon/3$, it follows that $f'(B)<-2\varepsilon/3$. Similarly, if we consider any $C>B$ such that $f(x)\geq2\varepsilon/3$ throughout the interval $[B,C]$, then throughout this interval we also have $f'(x)<-\varepsilon/3$. By the mean value theorem, there is a point $y\in[B,C]$ such that 
$$
\frac{fC)-f(B)}{C-B}=f'(y)<\frac{-\varepsilon}3.
$$
A bit of algebraic manipulation (which I might have done correctly) converts this  (plus the information that $f(C)\geq2\varepsilon/3$) to 
$$
C<B+\frac3\varepsilon f(B)-2.
$$
Even if my algebra wasn't perfect, we get some upper bound on $C$. That is, $f$ cannot remain $\geq2\varepsilon/3$ from $B$ all the way out to $\infty$. 
For the rest of the proof, fix $C$ as the first point after $B$ where $f$ takes a value $\leq2\varepsilon/3$. (There is a first such point because $f$, being differentiable,is continuous.) Continuity of $f$ also gives us that $f(C)=2\varepsilon/3$.  To complete the proof, it suffices to show that $f(x)<\varepsilon$ for all $x>C$. So suppose not, and let $D$ be the first counterexample. Again, continuity of $f$ implies that there is a first such $D$ and that $f(D)=\varepsilon$. 
For all $x\in[C,D]$, we have, by our choice of $D$, that $f(x)<\varepsilon=f(D)$. Therefore, for $x\in[C,D)$, 
$$
\frac{f(D)-f(x)}{D-x}>0
$$
and, since $f$ is differentiable at $D$, $f'(D)\geq0$. But then $f(D)+f'(D)\geq f(D)=\varepsilon$, contrary to the fact that, since $D>A$, our choice of $A$ ensures $f(D)+f'(D)<\varepsilon/3$.
Therefore, no such $D$ can exist, and the proof is complete.
A: The most elegant proof is based on L'Hôpital's rule by writing $f(x) = \dfrac{e^xf(x)}{e^x}$. See the answers to the duplicate questions mentioned in the comments plus  YuiTo Cheng's comment to my answer.
Here is a proof not using L'Hôpital's rule.
It suffices to consider $L = 1$: Define $g(x) = f(x) + 1 - L$; then $g(x) + g'(x) \to 1$. If we can show that $g(x) \to 1$, then $f(x) = g(x) - 1 + L \to L$.
Let us show that for each $\epsilon \in (0,L)$ one has $f(x) > 1 - \epsilon$ for $x  > R(\epsilon)$. Similarly can can show that $f(x) < 1 + \epsilon$ for $x  > R'(\epsilon)$. This will prove that $f(x) \to 1$ as $x \to \infty$.
Since $f(x) + f'(x) \to 1$, we find $r$ such that $f(x) + f'(x) > 1 -\epsilon/2$ for $x > r$.


*

*There exists $R \ge r$ such that $f(R) >  1 - \epsilon$. Assume $f(x) \le  1 - \epsilon$ for all $x \ge r$. Then $f'(x) > 1 - \epsilon/2 - f(x) \ge 1 - \epsilon/2 -(1 - \epsilon) = \epsilon/2$ for all $x \ge r$. Hence $f(x) = f(r) + \int_r^x f'(t)dt > f(r) + (x - r)\epsilon/2 > 1 - \epsilon$ for suffiently large $x$, a contradiction.

*$f(x) > 1 - \epsilon$ for all $x > R$. Assume there exists $x_1 > x$ such that $f(x_1) \le 1 - \epsilon$. Let $\xi = \inf \{ x_1 > R \mid f(x_1) \le 1 - \epsilon \}$. Then $f(\xi) = 1 - \epsilon$ (note that if $f(\xi) < 1 - \epsilon$, then $f(x_1) < 1 - \epsilon$ for $\lvert x_1 - \xi \rvert < \delta$) and $f(x) > 1 - \epsilon$ for $x \in [R,\xi)$. We have $f'(\xi) > 1 - \epsilon/2 - f(\xi) = \epsilon/2$, thus $f'(x) > 0$ for $\lvert x -\xi \rvert < \delta$. Hence $f$ is strictly increasing on $(\xi - \delta, \xi + \delta)$. Pick $x \in [R,\xi) \cap (\xi - \delta, \xi + \delta)$. Then $f(\xi) > f(x) > 1 - \epsilon$, a contradiction.
A: Consider $g(x)=e^xf(x)$. Then by the extended mean value theorem for any pair $a<b$ there is some $c\in(a,b)$ with
$$
\frac{g(b)-g(a)}{e^b-e^a}=\frac{g'(c)}{e^c}=f(c)+f'(c)\tag1
$$
By the stated convergence, there is for some fixed $0<ε<1$ an $x_ε$ so that $$f(x)+f'(x)\in(L-ε,L+ε) ~~\text{ for all } ~~ x>x_ε\tag2.$$ Then for any pair $b>a>x_ε$ by (1) and (2)
$$
(1-e^{a-b})(L-ε)<f(b)-e^{a-b}f(a)<(1-e^{a-b})(L+ε)\tag3
$$
Now select $b_ε$ for fixed $a$ so that $\max(1,L,|f(a)|)e^{a-b_ε}<ε<1$. Then for any $b>b_ε$
$$
 L-4ε<f(b)<L+4ε.\tag4
$$
This proves that $f(x)\xrightarrow{x\to\infty}L$ and in consequence $f'(x)\xrightarrow{x\to\infty}0$.
