Circular reasoning in L'Hopital's rule Suppose we have a function $f(x)$ that satisfies:
$$\lim_{x\to\infty}f(x)=L$$
Where $L\in\mathbb{R}$. Is this true?
$$\lim_{x\to\infty}f'(x)=0$$
My approach was simply this:
$$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{xf(x)}{x}=L$$
And applying L'Hospital's rule we have:
$$\lim_{x\to\infty}\frac{xf(x)}{x}=\lim_{x\to\infty}\frac{f(x)+xf'(x)}{1}=L$$
$$\lim_{x\to\infty}f(x)+xf'(x)=L+\lim_{x\to\infty}xf'(x)=L$$
And finally:
$$\lim_{x\to\infty}xf'(x)=0$$
Now, the only way this is possible is if $\lim_{x\to\infty}f'(x)\neq\infty$ and $\lim_{x\to\infty}f'(x)\neq A\in\mathbb{R}$ , because otherways the $\lim_{x\to\infty}xf'(x)$ would go to infinity. In conclusion, $\lim_{x\to\infty}f'(x)=0$
Is this in any way circular reasoning? I'm especially worried about the part when we apply the L'Hospital's rule.
 A: (Paraphrased from Wikipedia.)
L'Hôpital's rule:
Given functions $f$ and $g$ which are differentiable on an open interval $I$, except possibly at a point $c \in I$, if 
$$ \lim _{x \to c}F(x)=\lim _{x\to c}G(x)=0 \text{ or }\pm \infty, \tag{1.} $$
$$ G'(x)\neq 0 \text{ for all }x \in I, \text{ with }x \ne c, \text{ and} \tag{2.} $$
$$ \lim_{x \to c}\frac{F'(x)}{G'(x)} \text{ exists.} \tag{3.} $$
then
$$\lim_{x \to c} \frac{F(x)}{G(x)} =\lim_{x \to c} \frac{F'(x)}{G'(x)}. \tag{4.}$$
You used $F(x) = xf(x)$ and $G(x) = x$ and $I = (x_0, \infty)$ for some $x_0 < 0$. 
Since $\lim _{x\to \infty}G(x)= \infty$, condition $(1.)$ requires that
$$\lim _{x \to \infty}xf(x) = \infty. \tag{A.}$$
Condition $(2.)$ is satisfied by $G(x)=x$.
Condition $(3.)$ requires that 
$$\lim_{x \to \infty}[f(x)+xf'(x)] \text{ exists.} \tag{B.}$$
If conditions $(A.)$ and $(B.)$ are met, then, by L'Hôpital's rule,
$$ \lim_{x \to \infty} f(x) = \lim_{x \to \infty}[f(x)+xf'(x)]$$
Others have shown you that counter examples do exists.
A: Suppose that $f(x)=\dfrac{\sin(x^2)}x$. Then $\lim_{x\to\infty}f(x)=0$, but the limit $\lim_{x\to\infty}f'(x)$ doesn't exist.
If you try to apply L'Hopital's Rule here as you did, you will be working with$$\lim_{x\to\infty}\frac{x\sin(x^2)}{x^2}.$$But if $g(x)=x\sin(x^2)$, then the limit $\lim_{x\to\infty}g'(x)$ doesn't exist. Therefore, you cannot apply L'Hopital's Rule here.
