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)?
Best Answer
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.