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


  • $\begingroup$ Related: math.stackexchange.com/q/407654 $\endgroup$ – Jonas Meyer Oct 8 '13 at 3:17
  • $\begingroup$ @JonasMeyer, your link is same my solution,But my problem have other problem,But Thank you all the same $\endgroup$ – user94270 Oct 8 '13 at 3:19
  • $\begingroup$ nanchangjian: Do you mean to hypothesize that $f'$ is also continuous? Otherwise this is false, because you can find differentiable functions $f$ satisfying the hypotheses with $f'$ discontinuous. If $f'$ is continuous, then you may also refer to math.stackexchange.com/questions/75491/…. $\endgroup$ – Jonas Meyer Oct 8 '13 at 3:23
  • $\begingroup$ Sorry,I'm very sorry,Now I have edit:becasuse this problem from chinese say:f(x)是连续可导 $\endgroup$ – user94270 Oct 8 '13 at 3:31
  • $\begingroup$ nanchangjian: I don't know Chinese, but according to Google Translate the translation to English is "continuously differentiable," which means $f'$ is continuous. Thank you. $\endgroup$ – Jonas Meyer Oct 8 '13 at 3:34

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.

  • $\begingroup$ @JonasMeyer,yes,I agree with you,Thank you, $\endgroup$ – user94270 Oct 8 '13 at 3:23
  • $\begingroup$ and dear @N.s.,can you post your all solution? Thank you $\endgroup$ – user94270 Oct 8 '13 at 3:24
  • $\begingroup$ @JonasMeyer Yea missed that. $\endgroup$ – N. S. Oct 8 '13 at 3:27
  • $\begingroup$ sorry,I have edit.Now can you post your all solution? Thank you $\endgroup$ – user94270 Oct 8 '13 at 3:32

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