If $\lim_{t\rightarrow \infty}f(t) = 0$, then is it true that $\lim_{t\rightarrow\infty}f'(t) = 0$

Question

Suppose that $f:\mathbb{R}\rightarrow\mathbb{R}^n$ is differentiable and that $\lim_{t\rightarrow \infty}f(t) = 0$. Does it follow that $\lim_{t\rightarrow\infty}f'(t) = 0$?

Edit: there are now several great counter examples. No need for new answers. Thanks guys!

Answer 1

This is not true; basically, we could tend to zero, but still have very sharp oscillation at infinity. Just take $f(x) = \frac{sin(x^2)}{x}$. Then $|f(x)|$ tends to zero as $x \to \infty$, but taking the derivative, we see that the derivative does not.

Answer 2

No. Let $f(x)=\frac1x \sin (x^2)$. Just because a function is close to zero, that does not imply its slope is close to flat.

Answer 3

Counter example: $$f(t) = \frac{1}{t} \sin t^2 $$ as $$ f'(t) = 2 \cos t^2 - \frac{1}{t^2} \sin t^2 $$ which does not converge as $t \rightarrow \infty$.

Answer 4

Nope. $$ f(x)=\sum_{n\geq 1}\frac{1}{n} e^{-n^2(x-n)^2} $$ is a $C^1(\mathbb{R})$ function, but while $f(x)\to 0$ as $x\to +\infty$ (we actually have $f(x)\ll\frac{1}{x}$) the limit $\lim_{x\to +\infty}f'(x)$ does not exist, and the following diagram representing the graph of $f(x)$ over $(0,11)$ should help you in understanding why:

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