I am attempting to prove that if $a_n$ is a bounded sequence of real numbers then
$$\lim_{x\to1^-}(1-x)\left[\frac{d}{dx}(1-x)\sum_{n=1}^{\infty}a_nx^n\right]=0$$
My approach is to first make some algebraic manipulations, namely we see that
\begin{align*} 1&=\lim_{x\to1^-}\frac{(1-x)\sum_{n=1}^{\infty}a_nx^n}{(1-x)\sum_{n=1}^{\infty}a_nx^n}\\ &=\lim_{x\to1^-}\frac{1}{(1-x)\sum_{n=1}^{\infty}a_nx^n}\left(\frac{1-x}{\frac{1}{\sum_{n=1}^{\infty}a_nx^n}}\right)\\ \end{align*}
The reason I want to do this is that if I were able to apply L'Hopital's rule to
$$\frac{1-x}{\frac{1}{\sum_{n=1}^{\infty}a_nx^n}}$$
then I would get that
\begin{align*} 1&=\lim_{x\to1^-}\frac{1}{(1-x)\sum_{n=1}^{\infty}a_nx^n}\left(\frac{-1}{-\frac{\sum_{n=1}^{\infty}na_nx^{n-1}}{\left(\sum_{n=1}^{\infty}a_nx^n\right)^2}}\right)\\ &=\lim_{x\to1^-}\frac{\sum_{n=1}^{\infty}a_nx^n}{(1-x)\sum_{n=1}^{\infty}na_nx^{n-1}}\\ \end{align*}
From there we can subtract $1$ from both sides and multiply top and bottom by $(1-x)$ to get that
$$\lim_{x\to1^-}\frac{\left(1-x\right)\sum_{n=1}^{\infty}a_{n}x^{n}-\left(1-x\right)^2\sum_{n=1}^{\infty}na_{n}x^{n-1}}{\left(1-x\right)^2\sum_{n=1}^{\infty}na_{n}x^{n-1}}=0$$
Since
$$\left(1-x\right)^2\sum_{n=1}^{\infty}na_{n}x^{n-1}$$
is bounded, the only way for this quantity to go to zero would be for
$$\left(1-x\right)\sum_{n=1}^{\infty}a_{n}x^{n}-\left(1-x\right)^2\sum_{n=1}^{\infty}na_{n}x^{n-1}=(1-x)\left[\frac{d}{dx}(1-x)\sum_{n=1}^{\infty}a_nx^n\right]$$
to go to $0$, thus yielding what we want.
I am not sure if this use of L'Hopitals is (or can be) justified, since the limit of $$\frac{-1}{-\frac{\sum_{n=1}^{\infty}na_nx^{n-1}}{\left(\sum_{n=1}^{\infty}a_nx^n\right)^2}}$$ as $x\to1^-$ is not required to exist. Is there any way I can make this argument rigorous?
EDIT: If I had the pair of inequalities
$$\limsup_{x\to 1^-}k(x)\frac{f(x)}{g(x)}\leq \limsup_{x\to 1^-}k(x)\frac{f'(x)}{g'(x)}$$
$$\liminf_{x\to 1^-}k(x)\frac{f'(x)}{g'(x)} \leq \liminf_{x\to 1^-}k(x)\frac{f(x)}{g(x)}$$
for differentiable functions $f$, $g$ and $k$ on $[0,1)$ then I could resolve my issue. On wikipedia it states that
$$\liminf_{x\to1^-}\frac{f'(x)}{g'(x)}\leq \liminf_{x\to1^-}\frac{f(x)}{g(x)} \leq \limsup_{x\to1^-}\frac{f(x)}{g(x)}\leq \limsup_{x\to1^-}\frac{f'(x)}{g'(x)}$$
but I can't complete the argument for when the factor of $k(x)$ is added.
from Hot Weekly Questions - Mathematics Stack Exchange
Milo Moses
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