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

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 $n$.