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Suppose $f : [0, ∞) → R $ has continuous first and second derivatives and $f(x) → 0$ as $x → ∞$. If $f'(x) → b $ as $ x → ∞$, show that $b = 0.$

I tried using L'Hopital Rule here, by constructing $\lim_{x\to \infty}\frac{f(x)}{f'(x)}=$$\lim_{x\to \infty}\frac{f'(x)}{f''(x)}$. However, I 'm not sure if L'Hopital Rule can be used here because $\lim_{x\to \infty}f'(x)=0$ can not be used as condition.

I'm stuck in this way, and can't see other ways to solve this problem. Any hint?

XXWANGL
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2 Answers2

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Hint: for any $[a,b]$, $\exists \theta(a,b): \theta \in [a,b], f'(\theta)={f(b)-f(a) \over b-a}$. Now take $a_n=f'(\theta(n,n+1))$, it's obvious that $lim_{x \to +\infty}a_n = b$. Show that $lim_{n \to \infty}a_n = 0$.

Note that the proof holds for any $f_0 : f(x) \to f_0$ as $x \to +\infty$.

Abstraction
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  • I don't get what you are trying to do here. Could you plz be more specific? – XXWANGL Dec 07 '15 at 15:05
  • I offer you some specific sequence $a_n$ so that, on one hand, $\lim_{n \to \infty} a_n = b$ and on the other, $\lim_{n \to \infty} a_n = 0$. – Abstraction Dec 07 '15 at 15:14
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Since $$f'(x)= \lim_{h\to0} \frac{f(x+h)-f(x)}{h},$$ we have, replacing $h$ by $f(x)$ and taking $x\to+\infty$: $$ \lim_{x\to+\infty} f'(x)= \lim_{x\to+\infty} \frac{f(x+f(x))}{f(x)} - 1,$$ and thus $$ \lim_{x\to+\infty} \frac{f(x+f(x))}{f(x)} =1+b. \tag{$\star$} $$ Now, $b>0$ would imply $f(x)<0$ for large enough $x$, whence $$ \frac{f(x+f(x))}{f(x)} <1, $$ for those large $x$, contradicting $(\star)$. A contradiction is analogously derived from $b<0$. Hence, $b=0$.

Vincenzo Oliva
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