Prove for every polynomial $p(x)$ of degree less or equal than $n$ satisfying $\lim_{x\to x_0} \frac{p(x)}{(x-x_0)^n}=0$ that $p\equiv 0$ [closed]

For every polynomial $p(x)$ of degree less or equal than $n$ satisfying$$\lim_{x \to x_0} \frac{p(x)}{(x - x_0)^n} = 0, $$we have $p \equiv 0$.

This problem is from my notes booklet on Taylor and it is used to prove a theorem but I'll just put a typical tag (you can add any). I completed the induction step which is missing from my notes.

By induction, we start for $n=1$ we can consider a 1st degree polynomial $p(x) = a(x-x_0)+b$. Thenfor every $x\neq x_0$:$$\frac{p(x)}{x - x_0} = a + \frac{b}{x - x_0}. \tag{1}$$We see that$$p(x_0) = \lim_{x \to x_0} p(x) = \lim_{x \to x_0} \frac{p(x)}{(x - x_0)^n}(x - x_0)^n = 0. \tag{2}$$

We fix $x \to x_0$ and we get$$0 = a + \frac{b}{x - x_0} = a + 0 = a, $$since $p(x_0) = b = 0$. Therefore, $a = b = 0 \implies p(x) \equiv 0$. If $p(x)$ is a polynomial of degree $ \leq n \implies p(x) = (x - x_0) q(x)$, where $q(x)$ is a polynomial of degree $\leq n - 1$, $$\lim_{x \to x_0} \frac{q(x)}{(x - x_0)^{n-1}} = \lim_{x \to x_0} \frac{p(x)}{(x - x_0)^n} = 0. $$Then, $q(x) = 0 \implies p(x) \equiv 0$.