How to calculate infinite limit for function and its derivative? 
How to do this question?
Tried separating the limit to
$$\lim_{x \rightarrow +\infty} f(x) + 2 \lim_{x \rightarrow +\infty} f'(x) + \lim_{x \rightarrow +\infty} f"(x) = k$$
but seems not working
 A: Let
$$F(x)=e^xf(x)\quad\text{and}\quad G(x)=e^x.$$The hypothesis rewrites$$\lim_{+\infty}\frac{F''}{G''}=k.$$By L'Hôpital's rule (used twice), this implies $$\lim_{+\infty}\frac{F'}{G'}=k\quad\text{and}\quad\lim_{+\infty}\frac FG=k,$$i.e. $\lim_{+\infty}(f+f')=k$ and $\lim_{+\infty}f=k$, whence the claim. Note that the hypothesis of non-nullity of $\lim_{+\infty}f$, nor even of its existence, was useless.
A: We first prove this lemma:
Lemma: let h be a difrrentiable function in $(a,\infty)$ such that $\lim_{x \rightarrow +\infty} h(x) + h^{'}(x) =k$ thus $\lim_{x \rightarrow +\infty} h(x) = k$ and $\lim_{x \rightarrow +\infty} h^{'}(x) = 0$.
It's enough to prove that $lim_{x \to \infty}h(x)=k$.
Let $\epsilon > 0$ let's assume by contradiction that for all $N>0$ there exit $x>N$ such that $h(x) > k + \epsilon$. because $\lim_{x \to +\infty} h(x) + h^{'}(x) =k$   there exist $M>0$ such that for all $x>M$ we have $k-\frac{\epsilon}{2} < h(x)+h^{'}(x) < k-\frac{\epsilon}{2}$.
Notice that there exist $y>M$ such that $h(y)  <k + \epsilon$ otherwise for all $y>M$ $h^{'}(y)\le k-\frac{\epsilon}{2} - h(y) \le k-\frac{\epsilon}{2} -(k + \epsilon)= -\frac{\epsilon}{2}$. and thus $\lim_{y \to \infty} f(y) = -\infty$ (from lagrange).
So let $y>M$ such that $h(y) < k+ \epsilon$ from the assumption there exist $x>y$ such that $h(x) > k +\epsilon$. From continouty of $h$ there exist $a \in (y,x)$ such that $h(a) = k + \epsilon$. Define $z = \sup\{b\in (y,x) | h(b) =k+ \epsilon\}$ notice that z is well defined as $a \in  \{b\in (y,x) | h(b) =k+ \epsilon\}$. Also from continouty $h(z)=k + \epsilon$ and $z<x$. And finally notice that for every $c \in (z,x)$ $h(c) > k +\epsilon$(othewise there will be a point $p$ between $(z,x)$ such that $h(p)=k + \epsilon$ in cotrandiction to that that $z$ is the suprimum.
notice that $h(x)>h(z)$ and thus from lagrange there exist a point $b \in (z,x)$ such that $h'(b) > 0$. But this is a contradiction as for every point $c \in (z,x)$ $h^{'}(c)\le k-\frac{\epsilon}{2} - h(c) \le k-\frac{\epsilon}{2} -(k + \epsilon)= -\frac{\epsilon}{2}<0$
And thus there exist $N_1>0$ such that for every $x>N_1$ we have $h(x) \le k+ \epsilon$. In the same way we can prove that there exist $N_2>0$ such that for every $x>N_2$ we have $h(x) \ge k - \epsilon$. And thus for $N = \max\{N_1,N_2\}$ for all $x > N$ we have $|h(x) -k| \le \epsilon$.
And thus $lim_{x \to \infty}h(x)=k$. So we get the lemma.
Now define $h(x) = f(x) +f^{'}(x)$. Thus the assumtion in the question tells us that $\lim_{x \rightarrow +\infty} h(x) + h^{'}(x) =k$.And thus from the lemma we get that  $\lim_{x \rightarrow +\infty} h(x) = k$ and $\lim_{x \rightarrow +\infty} h^{'}(x) = 0$.
Now $k = \lim_{x \rightarrow +\infty} h(x) = \lim_{x \rightarrow +\infty} f(x) + f'(x)$. Abd again from the lemma we get that $\lim_{x \rightarrow +\infty} f(x) = k$ and $\lim_{x \rightarrow +\infty} f^{'}(x) = 0$.
And now from limit arritmetic we can get that
$\lim_{x \rightarrow +\infty} f^{''}(x) = \lim_{x \rightarrow +\infty} (f^{''}(x) +2f^{'}(x)+f(x))-2f^{'}(x)-f(x)=k+2\cdot 0 - k = 0$
