TITLE: How prove this $\lim_{x\to+\infty}(f'(x)+f(x))=l$
QUESTION [6 upvotes]: let $f(x)$ is continous and $f'(x)$ is continous  on $[0,\infty)$,show that

$$\lim_{x\to+\infty}(f'(x)+f(x))=l$$
  if and only if: $\displaystyle\lim_{x\to+\infty}f(x)=l$ and $f'(x)$ is uniformly continuous on $[0,+\infty)$.

How prove this it? Thank you.

I can prove this if $$\lim_{x\to+\infty}(f'(x)+f(x))=l$$
  then we have  $\displaystyle\lim_{x\to+\infty}f(x)=l$

My Part of the
Solution:
without loss of we let $l=0$,Give $\epsilon>0$,let $a>0$ be such that $|f(x)+f'(x)|<\epsilon$ for $x\ge a$
Then by the generalized mean value theorem there is $\xi\in (a,x)$ such that
$$\dfrac{e^x f(x)-e^af(a)}{e^x-e^a}=f(\xi)+f'(\xi)$$
Thus
$$|f(x)-f(a)e^{a-x}|<\epsilon|1-e^{a-x}|$$
so
$$|f(x)|<|f(a)|e^{a-x}+\epsilon|1-e^{a-x}|$$
so
$$|f(x)|<2\epsilon$$ for sufficiently large $x$.
But How can prove $f'(x)$ is uniformly continuous on $[0,+\infty)$
and other question: How prove if$\displaystyle\lim_{x\to+\infty}f(x)=l$ and $f'(x)$ is uniformly continuous on $[0,+\infty)$
then we have 

$$\lim_{x\to+\infty}(f'(x)+f(x))=l$$

REPLY [2 votes]: Hint 1 If $\lim_{x \to \infty} f(x)=l$ then $\lim_{x \to \infty} f'(x)=0$.
Thus, for each $\epsilon >0$ there exists an $a$ so that $|f'(x)|< \epsilon$ for $x \in (a, \infty)$. You also know that $f'(x)$ is uniformly continuous on $[0,a]$ (why)?
Hint 2 For the other question, try to apply the generalized mean value Theorem for $(x,x+a)$ when $a$ is very small and $x$ very large.