Let $q$ be a polynomial with $\deg(q) \le n$. Show if $q=o(x^n)$ for $x \to 0$ then $q$ is the zero polynomial

Question

Problem

Let $q$ be a polynomial with $\deg(q) \le n$. Show if $q=o(x^n)$ for $x \to 0$ then $q$ is the zero polynomial.

One must use proof by induction.

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I have used two hours on just understanding the question. I do not think I can come to a solution without help. I welcome any help. I also have a couple of questions

Q1: The zero polynomial is the polynomial with all the coefficients being zero, right? So $q = 0$?

Q2: I do I start the induction proof in this case?

Edit: it's only allowed to use the basic theory about limits and continuity.

Answer 1

Induction doesn't seem to be a good way to prove that.

I would use contradiction instead: suppose $m:=\text{deg}(q)>0$, then you can apply l'Hôpital's rule for $m$ times to the limit $$ \lim_{x\to0}\frac{q(x)}{x^n}, $$ until you get $$\lim_{x\to0}\frac{c}{x^{n-m}}, $$for some positive constant $c$. As $m\leq n$, this limit is non-zero. That's the contradiction, as $q(x)=o(x^n)$ by hypothesis.