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Suppose $f(t)$ is continuously differentiable and $c$ is a finite constant. We know that \begin{equation} \lim\limits_{t \to \infty} f(t)=c \implies\lim\limits_{t \to \infty} f'(t) = 0 \quad \text{is NOT true!}. \end{equation}

According to Barbalat's Lemma, this is true if $f'(t)$ is uniformly continuous. But can anyone give me a counterexample showing that $\lim\limits_{t \to \infty} f(t)=c$ NOT implying $\lim\limits_{t \to \infty} f'(t) = 0 $?

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    $\begingroup$ If $f$ has no derivative then the first limit does not even exist. Please provide some more context and hypotheses. Please do give us your views on the problem as well. $\endgroup$ – Bill O'Haran Dec 18 '18 at 11:14
  • $\begingroup$ @BillO'Haran Please see the edit. You are right, but you can think further, considering that $f(t)$ is differentiable. However, i am sorry for missing this condition. $\endgroup$ – winston Dec 18 '18 at 13:38
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I take that $f'$ should exist. An interesting example is if:

$$f(x) = \int_0^x \sin(t^2) \mathrm dt$$

Then: $$\lim_{x \to \infty} f(x) = \frac 1 2 \sqrt{\frac \pi 2}$$

but: $$\lim_{x \to \infty} f'(x) = \lim_{x \to \infty} \sin(x^2)$$

which doesn't exist.

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  • $\begingroup$ +1 If the limit of $f'$ exists, though, then it is $0$ $\endgroup$ – Ant Dec 18 '18 at 11:37
  • $\begingroup$ Why is the limit $1/2 *\sqrt{\pi/2}$? $\endgroup$ – winston Dec 18 '18 at 15:48
  • $\begingroup$ @winston Well, $\lim_{x \to \infty} f(x) = \int_0^\infty \sin (t^2) \mathrm dt$. There are many, many ways to then evaluate this. You can find some here and here. If you are just interested in showing that the integral converges, this will be more of interest. $\endgroup$ – George Coote Dec 18 '18 at 15:56
  • $\begingroup$ @GeorgeCoote Does this imply that $\sin(x^2)$ is not uniformly continuous? $\endgroup$ – winston Dec 18 '18 at 16:06
  • $\begingroup$ @winston: you have guessed correctly that $\sin x^2$ is not uniformly continuous. Why? Because $x^2$ is not uniformly continuous. $\endgroup$ – Paramanand Singh Dec 18 '18 at 16:50
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$$f(t)=\frac{\sin t^2}{t}.$$ Derivative $[t \cos(t^2)\cdot 2t-\sin(t^2)]/[t^2]$ is $2 \cos(t^2)$ plus something going to zero.

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