Proof of L'Hospital with power series

Question

I'm having a bit of problem with this question. I feel like I have to prove the l'hospital's rule but I don't know where to start especially because I have to use the power series.

Suppose that the power series $\sum_{n=0}^\infty a_nx^n$ and $\sum_{n=0}^\infty b_nx^n$ converge in some non-trivial interval centered at zero, and consider functions $f(x)=\sum_{n=0}^\infty a_nx^n$ and $g(x)=\sum_{n=0}^\infty b_nx^n$. Prove (without using L'Hospital's Rule!) that if $\displaystyle\lim_{x\to 0}f(x)=\lim_{x\to 0}g(x) = 0$, then

$$\lim_{x\to 0}\frac{f(x)}{g(x)} = \lim_{x\to 0}\frac{f'(x)}{g'(x)}.$$

Answer 1

Hint. First note, that $$ 0 = \lim_{x\to 0} f(x) = \lim_{x\to 0}\sum_n a_nx^n = a_0 $$ and likewise $b_0 = 0$.

Along the same line of thought, we have $$ a_1 = \lim_{x\to 0} f'(x), \quad b_1 = \lim_{x\to 0} g'(x) $$ Now $$ \frac{f(x)}{g(x)} = \frac{\sum_{n\ge 0} a_n x^n}{\sum_{n\ge 0} b_n x^n} = \frac{\sum_{n\ge 1} a_n x^n}{\sum_{n\ge 1} b_n x^n} = \frac{\sum_{n \ge 0} a_{n+1} x^n}{\sum_{n\ge 0} b_{n+1} x^n} $$ What happens for $x\to 0$ in the last fraction?

Answer 2

You may observe that, as $x \to 0$, $$ f(x)=a_1x+\sum_{n=2}^{\infty}a_nx^n $$ $$ g(x)=b_1x+\sum_{n=2}^{\infty}b_nx^n $$ giving $$ \frac{f(x)}{g(x)}=\frac{a_1+\sum_{n=2}^{\infty}a_nx^{n-1}}{b_1+\sum_{n=2}^{\infty}b_nx^{n-1}} \to \frac{a_1}{b_1} $$ Similarly, since $$ f'(x)=a_1+\sum_{n=2}^{\infty}na_nx^{n-1} $$ $$ g'(x)=b_1+\sum_{n=2}^{\infty}nb_nx^{n-1} $$ you get $$\frac{f'(x)}{g'(x)}=\frac{a_1+\sum_{n=2}^{\infty}na_nx^{n-1}}{b_1+\sum_{n=2}^{\infty}nb_nx^{n-1}} \to \frac{a_1}{b_1}$$ with $b_1 \neq 0$.