Set n as a natural number and suppose that P(x) = a0 + a1x + ... + apxp and Q(x) = b0 + b1x + ... + bqxq are polynomials of order p and q, respectively.
Such that $\lim_{x \rightarrow 0} (P(x)-Q(x))/x^n = 0$
Prove that aj = bj for all j = 0 ... n
I've tried using taylor series or l'hopitals rule but just can't make it work. Does anyone know how to do this?
Lemma. Let $R(x) = c_0 + c_1 x + \ldots + c_r x^r$ and assume that
$$\lim_{x \to 0} \frac{R(x)}{x^n} = 0$$
Then $c_j = 0$ for $j = 0, 1, \ldots, n$.
Proof. We do strong induction on $j = 0, 1, \ldots, n$: let $0 \leqslant j \leqslant n$ and suppose that for $i < j$ we have that $c_i = 0$. We're going to show that $c_j = 0$.
Note that
$$\lim_{x \to 0} \frac{R(x)}{x^j} = \lim_{x \to 0} \frac{R(x)}{x^n} \cdot x^{n-j} = 0$$
because $\frac{R(x)}{x^n} \to0$ and $x^{n-j}$ either tends to zero (if $j < n$) or is constantly equal to $1$ (if $j = n$).
On the other hand, since $c_i = 0$ for $i < j$:
$$\lim_{x \to 0} \frac{R(x)}{x^j} = \lim_{x \to 0} \frac{c_j x^j + c_{j+1} x^{j+1} + \ldots + c_r x^r}{x^j} = \lim_{x \to 0} c_j + c_{j+1} x + \ldots + c_r x^{r-j} = c_j,$$
hence $c_j = 0$. So the proof is complete. $\square$
Now to answer your question it suffices to take $R(x) = P(x) - Q(x)$ in the lemma above.