Prove that $ \lim_{x \to \infty} f^{n}(x) = 0$

Question

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 $\displaystyle \lim_{x \to \infty} f^{n}(x) = 0$?

That doesn't prove anything - L'Hopital's rule doesn't even apply here, since you don't know $f(x) \to 0$. — preferred_anon, Mar 20 '16 at 17:34
@DanielLittlewood $\lim_{x \to \infty} f(x)$ is a real number implies that $\lim_{x \to \infty} \dfrac{f(x)}{x^n} = 0$, so L'Hospital's rule does apply. — Puzzled417, Mar 20 '16 at 17:39
Nope, the condition for L'hopital is that the top and bottom tend to zero (or, equivalently, infinity). That might not be true. — preferred_anon, Mar 20 '16 at 17:44
@DanielLittlewood We have an infinite limit here though. Also I never assumed $f(x) \to 0$. — Puzzled417, Mar 20 '16 at 17:45
@DanielLittlewood What I am saying is this $\lim_{x \to \infty} (f(x) + x^n) = \infty$ and $\lim_{x \to \infty} x^n = \infty$. Thus we may use L'Hospitals. — Puzzled417, Mar 20 '16 at 17:48
You can easily modify what you started with. Replace $f(x)/x^n$ with $n!f(x)/x^n$. I agree that the first numerator tends to infinity. Having $f(x)\to r$, for $x$ sufficiently large, $0\le x^n-|r| \le |x^n+f(x)|$ — David P, Mar 20 '16 at 17:50
I have taken the freedom to clean-up your code a little, the line with formulae was so long that it dodn't display properly. Also, the parantheses were not needed. — Alex M., Mar 20 '16 at 17:57
The last equality appearts to be an error because L'Hopital doesn't apply. — zhw., Mar 20 '16 at 18:05
see related question http://math.stackexchange.com/q/730411/72031 — Paramanand Singh, Mar 21 '16 at 04:25

K.Power · Accepted Answer · 2016-03-20T18:46:47.630

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.

score 1 · Answer 2 · answered Mar 20 '16 at 18:20

1

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 $M=0.$

answered Mar 20 '16 at 18:20

zhw.

105,693

How do we know that $f'(x), f''(x),... $ are all differentiable? They may be differentiable nowhere. – Puzzled417 Mar 20 '16 at 19:38
You are assuming $f^{(n)}$ exists everywhere, right? – zhw. Mar 20 '16 at 20:00
For some $n$ we are. – Puzzled417 Mar 20 '16 at 20:03
Well you can't have a second derivative without having a first derivative, right? – zhw. Mar 20 '16 at 20:08
Right, and maybe the first derivative is defined nowhere. Recall that L'Hospital's rule for infinite limits is that the derivative is defined for all $x$ given some sufficiently large $N$ so that $x > N$. – Puzzled417 Mar 20 '16 at 20:16
1

Look, if $f^{(n)}$ exists on $(a,\infty)$ then so does $f^{(n-1)}$, hence so does $f^{(n-2)}$ etc – zhw. Mar 20 '16 at 20:33
Oh, I see. So we can be sure that the lower derivatives exist for some sufficiently large $N$. – Puzzled417 Mar 20 '16 at 20:33
This is the simpler and correct version of what OP is trying to do in his question. +1 – Paramanand Singh Mar 21 '16 at 04:28

Prove that $ \lim_{x \to \infty} f^{n}(x) = 0$

2 Answers2