Does existence of $\lim_{x \to 0} f(x)$ imply $\lim_{x \to 0} x f'(x) = 0$?

Question

Suppose we have a function $f : \mathbb{R}^+ \to \mathbb{R}$. It seems intuitive to me that if $\lim_{x \to 0} f(x)$ exists, then $\lim_{x \to 0} x f'(x) = 0$. I suspect that for real functions, there may be pathological counterexamples to this, but at least for analytic functions (where $0$ may be on the boundary of the analytic disk) then it should be true.

In the analytic case, I can give an argument for this that would convince a typical physicist like myself. $f(x)$ cannot have an essential singularity at $0$, because $\lim_{x \to 0} f(x)$ would not exist. So as $x \to 0$, it scales like $f(x) = c + O( x^\alpha )$ for some constant $c$ and $\alpha > 0$, since the limit exists. Therefore, $x f'(x) = O( x^\alpha ) \to 0$.

Two questions:

1. Is this true, and if so, how generally valid is it?

2. Is there a simpler proof that does not rely on scaling arguments? This seems like the kind of problem that one would typically address with a fairly general theorem like L'Hospital's rule. Another strategy is to write $xf'=(xf)'-f$, but this just shifts the burden of the problem to showing that $\lim_{x \to 0}[xf(x)]'=\lim_{x \to 0} f(x)$. I am guessing that the solution is totally obvious and I am just not seeing it.

Answer 1

You already suspect it may not be true in the real case. The misinterpretation of your question has already provided a few counterexamples which can only arise when the limit of $xf'(x)$ does not exist.

One thing to note in the real case is if $f$ is differentiable in an excluded neighborhood of $0$, $\lim_{x \to 0} f(x) = 0$ and $\lim_{x \to 0} x f'(x)$ exists, then it must be $0$. By L'Hopital's rule

$$0 = \lim_{x \to 0+} \frac{f(x)}{\log x} = \lim_{x \to 0+} xf'(x) , \\ 0 = \lim_{x \to 0-} \frac{f(x)}{\log (-x)} = \lim_{x \to 0-} -xf'(x) $$

The form of L'Hopital's rule that applies here is where $g(x) \to \infty$ implies that $\lim_{x \to 0} f(x)/g(x) = \lim_{x \to 0} f'(x)/g'(x)$ (when the limit on the RHS exists) without the assumption that $f(x) \to \infty$. Apparently this is not commonly known. See Case 2 in General Proof here to confirm.