# Prove that $\lim\limits_{x\to+\infty}f'(x)=\lim\limits_{x\to+\infty}f''(x)=0$

Given $$f \in C^3(\mathbb{R}).$$ Suppose that $$\lim\limits_{x\to+\infty}f(x)=a \in \mathbb{R}$$ and $$\lim\limits_{x\to+\infty}f'''(x)=0$$. Prove that $$\lim\limits_{x\to+\infty}f'(x)=\lim\limits_{x\to+\infty}f''(x)=0$$.

My solution: Since $$\lim\limits_{x\to+\infty}f(x)=a$$, there exists $$M>0$$ that for all $$x, x' \in \mathbb{R}$$, if $$x,x'>M$$ then $$f(x), f(x') \in \left(a-\varepsilon^2/2, a+\varepsilon^2/2\right)$$. This yields $$\vert f(x)-f(x') \vert < \varepsilon^2.$$

Therefore $$\frac{\vert f(x)-f(x') \vert}{\varepsilon} < \varepsilon$$. So, for all $$x>M$$, $$x'=x+\varepsilon$$ we have $$0\le\frac{\vert f(x+\varepsilon)-f(x) \vert}{\varepsilon} < \varepsilon.$$ Let $$\varepsilon \to 0$$ we have $$0 \le f'(x) \le 0 \text{ or } f'(x)=0.$$

That means $$\lim\limits_{x\to+\infty}f'(x)=0$$.

Doing similarly as above we have $$\lim\limits_{x\to+\infty}f''(x)=0.$$

My question: $$\lim\limits_{x\to+\infty}f'''(x)=0$$ is really necessary or not (because I don't use it in my proof)?

• Jan 18, 2021 at 17:52
• See the generalization at the end: math.stackexchange.com/q/730411/72031 Jan 19, 2021 at 0:49
• You need $f'''$ to be bounded and not necessarily tending to $0$. Jan 19, 2021 at 0:58
• Also see the handling of general result : math.stackexchange.com/a/731013/72031 Jan 19, 2021 at 0:59
• Your proof is not correct. If you wish to show that $|(f(x+h) - f(x)) /h|<\epsilon$ ensure that $h$ is independent of $\epsilon$. Jan 19, 2021 at 1:09

First, consider $$f(x)=\frac{\sin(x^2)}{x}.$$ As $$x\rightarrow\infty$$, this approaches $$0$$. On the other hand, $$f'(x)=\frac{2x^2\cos(x^2)-\sin(x^2)}{x^2}=2\cos(x^2)-\frac{\sin(x^2)}{x^2}$$ does not have a limit as $$x\rightarrow\infty$$. Therefore, the condition on $$f'''$$ is necessary for the result.

The problem with your setup is that you never specify when you're quantifying $$\varepsilon$$. You're actually quantifying it before computing $$M$$, so we'll write $$M=M(\varepsilon)$$. Next, you consider $$\frac{|f(x+\varepsilon)-f(x)|}{(x+\varepsilon)-x}.$$ At this point, you let $$\varepsilon$$ go to $$0$$, but here's your problem, as $$\varepsilon$$ goes to $$0$$, $$M(\varepsilon)$$ may go to infinity, so you don't get a bound on the derivative when $$\varepsilon$$ is too small, so the squeeze theorem cannot be applied.

Here's a proof that shows how the condition on the 3rd derivative could be utilized.

Take Taylor's remainder theorem to 3rd order and apply it to the intervals $$(x,x+h)$$, $$(x,x+k)$$, $$h\neq k$$. Now consider only the first interval since the second one can be done similarly.

There exists $$\xi\in(x,x+h)$$ such that $$f(x+h)=f(x)+f'(x)h+\frac{f''(x)}{2!}h^2+\frac{f'''(\xi)}{3!}h^3$$

Now since $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}f(x+h)=a$$, and $$\lim_{x\to\infty}f'''(\xi(x,h))=0$$ (why?) we conclude that $$\lim_{x\to\infty}\left(f(x+h)-f(x)-\frac{f'''(\xi)}{3!}h^3\right)=h\lim_{x\to\infty}\left(f'(x)+\frac{h}{2}f''(x)\right)=0 ~~~~~~~~~~(1)$$ Similarly we conclude $$\lim_{x\to\infty}\left(f'(x)+\frac{k}{2}f''(x)\right)=0~~~~~~~~~~~~~(2)$$

Taking the linear combinations $$(2)-(1)$$ and $$h(2)-k(1)$$ and exploiting the linearity of the limit operation when two limits exist, we obtain the desired result. The result is generalizable to a $$C^n$$ function whose $$n$$th derivative goes to zero for any $$n$$.

• Jan 18, 2021 at 19:11
• +1 Thanks to the limit of $f^{(n)}$ being $0$, your proof is simple. The result holds even when $f^{(n)}$ is bounded but the proof gets a bit complicated. See math.stackexchange.com/a/731013/72031 Jan 19, 2021 at 1:17
• There is a typo in the third derivative in the Taylor polynomial, also the denominator is 6, not 3.
– edm
Jan 19, 2021 at 5:55