For continuous $f(x)$, if $\lim_{x\to 0} f(x)/p(x)=0$ for every nonzero polynomial $p(x)$, then is the same true for nonzero analytic $p(x)$?

Question

Let $f(x)$ be a continuous real function on a neighborhood of 0 in $\mathbb{R}$.

Suppose $f(x)>0$ for $x\neq0$, and suppose $\lim_{x\to0} f(x)/p(x)$ for every nonzero polynomial $p(x)$.

My question is whether $\lim_{x\to 0} f(x)/P(x) = 0$ for any $P(x)$ analytic at 0, and not identically zero (on some connected neighborhood of 0)?

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My thinking so far is as follows:

1. It suffices to show $|P(x)|/f(x) \to \infty$ as $x\to 0$.

2. Similarly, by hypothesis, already $|p(x)|/f(x) \to \infty$ as $x\to 0$ for $p(x)$ any nonzero polynomial.

3. We are done if in the RHS of the expression $$ \lim_{x\to0} P(x)/f(x) = \lim_{x\to 0} \left( \frac{1}{f(x)} \lim_{n\to\infty} |P_n(x) | \right) = \lim_{x\to 0} \lim_{n\to\infty} \frac{|P_n(x) |}{f(x)} $$ we can interchange the two limits.

(Here $P_n(x)$ is the $n$th Taylor polynomial at $x=0$ of $P(x)$.)

Could anyone explain whether the changes (1) and (2) described in Step 3 above are justified? (Or, is my reasoning flawed?)

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Edit: removed a question about a trivial step (the 2nd equality above).

Answer 1

If $P$ is analytic near zero, then we can write $P(z) = z^k g(z)$ for some analytic $g$ with $g(0) \neq 0$. Then ${f(z) \over P(z)} = {f(z) \over z^k} {1 \over g(z)}$ and hence the limit is zero.