3
$\begingroup$

I was wondering if (1) $\lim_{x \to \infty} f(x) = a$, (2) $\lim_{x \to \infty} f'(x) = 0$, (3) $f(x) < a$ for all $x$ and

$$ (4) \; \; 0 \leq x ( a- f(x)) < C $$ for some positive constant $C$, then $\lim_{x \to \infty} x \; f'(x) = 0$.

Btw, the domain of $x$ is the nonnegative real numbers. Thanks,

$\endgroup$
2
  • 2
    $\begingroup$ I was wondering if you question wasn't in fact an exercise that was given to you, and if so, what you have tried already before posting it here. $\endgroup$
    – Olórin
    Feb 12, 2015 at 22:25
  • $\begingroup$ At least this question is plausible...$f(x)=a-e^{-x}$ could fit this profile... $\endgroup$
    – Troy Woo
    Feb 12, 2015 at 22:36

3 Answers 3

2
$\begingroup$

$$\text{If your limit exists then it must be zero, because we have:}$$

$$0\leq x(a-f(x))<C$$ $$\implies 0\leq \frac{a-f(x)}{1/x}<C$$ $$\implies 0\leq\lim_{x\to\infty}\frac{a-f(x)}{1/x}<C$$

$$\text{ But by L'hospital rule:}$$ $$\lim_{x\to\infty}\frac{a-f(x)}{1/x}=\lim_{x\to\infty}\frac{\frac{d}{dx}(a-f(x))}{\frac{d}{dx}1/x}=\lim_{x\to\infty}\frac{-f'(x)}{-1/x^2}=\lim_{x\to\infty}x^2f'(x)$$

$$\implies 0\leq \lim_{x\to\infty}x^2f'(x)<C$$ $$\implies \lim_{x\to\infty}xf'(x)=0$$

$\endgroup$
2
$\begingroup$

I cannot comment because of my reputation, but I wanted to point out that Ethan's solution is correct and twilight's counterexample is NOT a counterexample because:

1) $\lim_{x\to\infty}xf'(x)$ does not exist and,

2) $f(x) = a$ for $x = \frac{3\pi}{2} + 2\pi k,\ k \in \mathbb{N}$.

$\endgroup$
1
$\begingroup$

Sorry. I found a counterexample:

$$ f(x) = a - \frac{1+\sin x}{2+x} $$

So the fact that $x(a-f(x))$ is bounded above itself is not enough for the conjecture to hold true.

$\endgroup$
1
  • $\begingroup$ Note this is only true because $\lim_{x\to\infty}xf'(x)$ does not exist, this is because the cosine function appearing in $f'(x)$ constantly oscillates and prevents the limit from converging to a fixed value. However if his limit does exist, then it will always be zero. $\endgroup$ Feb 12, 2015 at 22:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .