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.

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.