TITLE: Prove that $ \lim_{x \to \infty} f^{n}(x) = 0$
QUESTION [8 upvotes]: Let $n$ be a positive integer. Assume $ \lim_{x \to \infty} f(x)$ and $ \lim_{x \to \infty} f^{(n)}(x)$ are both real numbers. Prove that $$ \lim_{x \to \infty} f^{n}(x) = 0$$  

We have that $$\lim_{x \to \infty} \frac{f(x)+x^n}{x^n} = \lim_{x \to \infty} \frac{f'(x)+nx^{n-1}}{nx^{n-1}} = \cdots = \lim_{x \to \infty} \frac{f^{(n)}(x)+n!}{n!} = \lim_{x \to \infty}f^{(n+1)}(x) = 1,$$ by L'Hospital's rule. But did I make a mistake or how does this show that $\displaystyle \lim_{x \to \infty} f^{n}(x) = 0$?

REPLY [2 votes]: Obviously as $\lim_{x\to \infty}f(x)$ exists we have$$\lim_{x \to \infty} \frac{f(x)+x^n}{x^n} =1$$
So your 2nd from last inequality is patently incorrect. You can't use L'Hospital's rule because you no longer have an indeterminate form. This is no issue though, as just equate your 3rd last step with your 1st. $$1=\lim_{x \to \infty} \frac{f^{(n)}(x)+n!}{n!}$$ so obviously from this $$1+\lim_{x \to \infty}\frac{f^{(n)}(x)}{n!}=1$$ and the result follows.