Prove that $ \lim_{x \to \infty} f^{n}(x) = 0$ 
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 $\displaystyle \lim_{x \to \infty} f^{n}(x) = 0$?
 A: 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.
A: For some reason the following version of L'Hopital is not as well known as the version stated for $\infty/\infty:$ Suppose $f,g$ are differentiable on some $(a,\infty)$ and $\lim_{x\to \infty} g(x) = \infty.$ If $\lim_{x\to \infty}f'(x)/g'(x) =L,$ then $\lim_{x\to \infty}f(x)/g(x) =L.$
This is the continuous version of its well known cousin, the Stolz-Cesaro theorem.
In the problem at hand, let $L = \lim f(x),M = \lim f^{(n)}(x) = M.$ Apply the above version of L'Hopital to get
$$\lim \frac{f(x)}{x^n} = \lim \frac{f'(x)}{nx^{n-1}} = \cdots = \lim \frac{f^{(n)}(x)}{n!} =\frac{M}{n!}.$$
Because $f(x) \to L,$ the limit of $f(x)/x^n$ is clearly $0.$ It follows that $M=0.$
