Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let f be a function that has a finite limit at infinity. It is true that this alone is not enough to show that its derivative converges to zero at infinity. So I was wondering weather there were any additional conditions for f that could give the desired outcome. I am also aware of Barbalat's Lemma but this requires uniform continuity, a property which in many occasions is not easy to verify. Thank you

share|cite|improve this question
4  
If its derivative's limit exists, then it should be zero. – Eastsun Nov 11 '12 at 10:40
    
What Eastsun says sounds right. For an example of a function with limit 0, but the derivative lacks a limit, see $\sin(x^2)/x.$ – Per Alexandersson Nov 11 '12 at 10:41

Without loss of generality, we have $f(x) \to 0$ as $x \to \infty$.

  1. If $\lim_{x \to \infty} f'$ exists, then it's clear that $f' \to 0$.
  2. If $f''$ is bounded, then $f' \to 0$. Morally, this is because $f$ must oscillate more and more tightly as $x \to \infty$ in order for $f' \not \to 0$, and $f''$ being bounded prevents that oscillation.
share|cite|improve this answer
    
The second part is covered here math.stackexchange.com/q/730411/72031 And +1 for adding this point. – Paramanand Singh Mar 23 at 4:07

In addition to the points covered in mixedmath's answer we can add the hypothesis that $f'$ is monotone. Monotonicity implies that $f'(x)$ either tends to $\infty$ or to $-\infty$ or to a limit $L$ as $x \to \infty$. Clearly since $f(x + 1) - f(x) = f'(c) \to 0$ with $x < x < x + 1$ it follows that the options for $f'$ to tend to $\pm\infty$ is not possible. Hence $f'(x) \to 0$ as $x \to \infty$.

The monotonicity of $f'$ can be guaranteed by assuming that $f''$ is of constant sign for all $x$ after a certain value.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.