Evaluating the limit of a quotient [Baby Rudin: 5.19] I will keep my explanation brief by separating it into three parts:
The Problem

Suppose $f$ is defined in $(-1,1)$ and $f'(0)$ exists. Suppose $-1<\alpha_n<\beta_n<1$, $\lim\limits_{n\to \infty}\alpha_n\to 0$, and $\lim\limits_{n\to \infty}\beta_n\to 0.$
Assume that $\alpha_n<0<\beta_n$.
Prove that
$\lim\limits_{n\to \infty} \frac{f(\beta_n)-f(\alpha_n)}{\beta_n-\alpha_n}=f'(0)$.

My Thoughts
I have been tackling this problem for a while, rolling it over in my head, and playing with the problem lead me to some ideas:

*

*Let $\gamma_n=\beta_n -\alpha_n$ then $\frac{f(\beta_n)-f(\alpha_n)}{\beta_n-\alpha_n}=\frac{f(\alpha_n +\gamma_n)-f(\alpha_n)}{\gamma_n}.$ Observing a similar version of this identity $\frac{f(\alpha_k+\gamma_n)-f(\alpha_k)}{\gamma_n}$, you would find that $\lim\limits_{n\to \infty} \lim\limits_{k\to \infty}\frac{f(\alpha_k+\gamma_n)-f(\alpha_k)}{\gamma_n}=f'(0).$


*Since $\lim\limits_{h\to 0} \frac{f(h)-f(0)}{h}=f'(0)$ then showing $\left|\frac{f(\beta_n)-f(\alpha_n)}{\beta_n-\alpha_n}-\frac{f(h)-f(0)}{h}\right|$ get closer together as $|\alpha_n|,|\beta_n|,|h|$ get small, would give a proof. Letting $h$ essentially mimic one of the given sequences reduces the problem a bit. Namely, letting $h(n)=\beta_n$ it follows that $$\left|\frac{f(\beta_n)-f(\alpha_n)}{\beta_n-\alpha_n}-\frac{f(h)-f(0)}{h}\right|<\left|\frac{f(\beta)-f(\alpha)}{\beta}-\frac{f(\beta_n)-f(0)}{\beta_n}\right|=\left|\frac{f(\alpha_n-f(0)}{\beta_n}\right|\textrm{.}$$
What I Want from You
I don't want the solution, I just want suggestions or tips that point me in the right direction.
Some questions to help guide what specific suggestion and tips might help:

*

*Would any of my ideas actually lead to anything fruitful and it if so any ideas on how to progress (like what should I focus on).

*Any parts of my proof that just doesn't seem right.

*Ideas or concepts that I am overlooking.


Also any suggestions how I can improve my questions readability is also greatly appreciated.
 A: You want to write that limit in terms of the usual definition of differentiation, thus
\begin{split}
&\left|\frac{f(\beta_n ) -f(\alpha_n)}{\beta_n - \alpha_n}-f'(0)\right| \\
&=\left| \frac{f(\beta_n) - f(0) + f(0) - f(\alpha_n)}{\beta_n -\alpha_n}-f'(0)\right|\\
&= \left| \frac{\beta_n }{\beta_n - \alpha_n}\left(\frac{f(\beta_n) - f(0)}{\beta_n}-f'(0)\right) -\frac{\alpha_n}{\beta_n -\alpha_n} \left(\frac{f(\alpha_n)-f(0)}{\alpha_n}-f'(0)\right)\right|\\
&\le \frac{\beta_n }{\beta_n - \alpha_n}\left|\frac{f(\beta_n) - f(0)}{\beta_n}-f'(0)\right| -\frac{\alpha_n}{\beta_n -\alpha_n} \left|\frac{f(\alpha_n)-f(0)}{\alpha_n}-f'(0)\right|
\end{split}
Note that we have used $\alpha_n<0< \beta_n$ in the last inequality. Now for any $\epsilon>0$, there is $N$ so that
\begin{align}
\left| \frac{f (\alpha _n )-f(0)}{\alpha_n}-f'(0)\right|<\epsilon
\end{align}
for all $n\ge N$ (and similar for $\beta_n$). Thus
$$\left| \frac{f(\beta _n)-f(\alpha_n)}{\beta_n-\alpha_n}-f'(0)\right|<\left(\frac{\beta_n }{\beta_n - \alpha_n} - \frac{\alpha_n}{\beta_n -\alpha_n}\right)\epsilon = \epsilon.$$
