If $f: \mathbb{R} \rightarrow \mathbb{R}$ is a differentiable function such that $\lim_{x\to\infty} f(x)$ exists and is finite, and $\lim_{x\to\infty} xf'(x)$ exists then find this limit
Intuitively I think $f'(x) \rightarrow 0$ as $ x \rightarrow \infty$ from first condition but that gives us a limit of $0ยท\infty$ form which can give us anything.
As $\ln(x)$ does not have a finite limit at infinity, and it's derivative is $\frac{1}{x}$, I suspect a function that does have a finite limit at infinity will have a derivative that decreases even faster such that $\frac{x}{\frac{1}{f'(x)}}$ will go to $0$ as $ x \rightarrow \infty$
Am I correct? If so, how can I formalize this as a proof?
Best Answer
Suppose $\lim_{x \to \infty} xf'(x)=c>0$. Then there exists $x_0$ such that $f(x)-f(x_0)=\int_{x_0}^{x} f'(t)dt>\int_{x_0}^{x} \frac c {2t}dt=\frac c 2[\ln x -\ln x_0]$ for all $x >x_0$. But this contradicts the hypothesis (since $\ln x -\ln x_0 \to \infty$). [Clearly, $\lim_{x \to \infty} xf'(x)=\infty$ also leads to a contradiction]. Similarly, $\lim_{x \to \infty} xf'(x)<0$ leads to a contradiction, so the limit must be $0$.