# If derivative goes to $0$, does the function have a limit?

Let $$f:\mathbb{R}\to\mathbb{R}$$ be a differentiable function. Assume that $$\lim_{x\to\infty}f'(x)=0$$, does that mean that $$\lim_{x\to\infty}f(x)$$ exists?

PS. What if we assume that $$f$$ is also bounded?

• By L'Hospital's Rule we get $f(x) /x\to 0$ and that's the best we can conclude here. – Paramanand Singh Dec 29 '18 at 1:37

Not necessarily. Consider $$f(x)=\sin(\ln(x))$$. Then $$f'(x)=\frac1x\cos(\ln(x))\to0$$, but $$f(x)$$ diverges, as it reaches $$-1$$ and $$1$$ infinitely often.

Not necessarily. See $$f(x)=\ln x$$:

$$\lim_{x\to \infty}f'(x)=\lim_{x\to\infty}\frac 1 x=0$$ $$\lim_{x\to\infty}f(x)=+\infty$$

• ah ok my bad. I missed this example. I wonder what if $f$ is also bounded. – David Lingard Dec 28 '18 at 23:00
• Maybe I should edit the question and add $f$ bounded. – David Lingard Dec 28 '18 at 23:01

No. Consider the function defined by cos(x) on [0,2pi], cos((x/2)-pi) on [2pi,6pi] and so on.

No.

Consider $$\lim_{t\to\infty}f'(t)=0$$

and

$$\lim_{x\to\infty}f(x)-f(a)=\lim_{x\to\infty}\int_{a}^{x}f'(t)dt$$ and $$a\geq0$$.

In this case, that integral converges if $$f(a)$$ exists for all $$a\geq0$$. for example the function $$f(x)=\ln x$$ doesn't satisfy that condition.

And if that integral converges, the limit exists.