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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 $?

winston
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    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. – Bill O'Haran Dec 18 '18 at 11:14
  • @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. – winston Dec 18 '18 at 13:38
  • Even if $f$ is differentiable, bounded, strictly increasing function, we can not conclude that $\lim_{t\to \infty} f'(t) = 0$. – River Li Jul 26 '19 at 12:09

3 Answers3

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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.

George C
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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.

coffeemath
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Just adding another counterexample here that isn't using trigonometric functions. It is conceptually very simple. Imagine a smoothened staircase where each step is equally steep but the height of each step is half the size of the one before it. The height converges to a limit but no matter how far you go there are still steep steps so the derivative can't go to 0. Now let's create that staircase:

Take the spike-looking function $\delta$ such that $\delta(x) = \max(0,1-|x|)$. This is a continuous function. Note that its integral over the real line is 1.

Now define $f(x)$ to be a function with such spikes at each natural number, but with the spike at $n$ squeezed horizontally by a factor of $2^n$. That is,

$f(x) = \sum_{n=1}^{\infty} \delta(2^n(x-n))$.

Its integral $g$ would have $$\lim_{x \to \infty} g(x) = g(0) + 1/2 + 1/4 + 1/8 + \cdots = g(0) + 1$$ but $$\lim_{x \to \infty} f(x)$$ does not exist.

SuhailSherif
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