# Limit at infinity of a bounded function

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,

• 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. Feb 12, 2015 at 22:25
• At least this question is plausible...$f(x)=a-e^{-x}$ could fit this profile... Feb 12, 2015 at 22:36

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

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

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

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

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 $x = \frac{3\pi}{2} + 2\pi k,\ k \in \mathbb{N}$.

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.

• 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. Feb 12, 2015 at 22:44