Show that the following goes to $f'(0)$ Let $f: (-1,1) \to R$ and $f'(0)$. Suppose that $-1 < a_n < 0 < b_n < 1$ and $a_n \to 0, b_n \to 0$ Show that
$$D_n = \frac{f(b_n) - f(a_n)}{b_n - a_n}$$ goes to $f'(0)$.
This is a specific case where we need to show that $f'(0)$ is the result, not a general point where the derivative is evaluated at.
I'm not quite sure where to start. I'm not sure what $f'(0)$ is, all I'm told is that it exists. So I'm trying to show this quantity goes to that, when I don't even know what $f'(0)$ is equal to. That is throwing me off. I'm also not sure the relation to $f'(0)$ that $D_n$ has. Can anyone clarify the question for me and maybe get me started? Thank you!
 A: Hint
\begin{align*}
\frac{f(b_n)-f(a_n)}{b_n-a_n}&=\underbrace{\frac{b_n}{b_n-a_n}}_{=1+\frac{a_n}{b_n-a_n}}\cdot \frac{f(b_n)-f(0)}{b_n}-\frac{a_n}{b_n-a_n}\cdot \frac{f(a_n)-f(0)}{a_n}\\
&=\frac{f(b_n)-f(0)}{b_n}+\frac{a_n}{b_n-a_n}\left(\frac{f(b_n)-f(0)}{b_n}-\frac{f(a_n)-f(0)}{a_n}\right)
\end{align*}
Since $\left|\frac{a_n}{b_n-a_n}\right|\leq 1$, taking $n\to \infty $ gives the wished result.
A: First write
$$
D_n  - f'(0) = \frac{{b_n }}{{b_n  - a_n }}\left( {\frac{{f(b_n ) - f(0)}}{{b_n }} - f'(0)} \right) \\ + \left( {1 - \frac{{b_n }}{{b_n  - a_n }}} \right)\left( {\frac{{f(a_n ) - f(0)}}{{a_n }} - f'(0)} \right).
$$
Using the fact that
$$
0 < \frac{{b_n }}{{b_n  - a_n }} < 1.
$$
we obtain
$$
\left| {D_n  - f'(0)} \right| \le \left| {\frac{{f(b_n ) - f(0)}}{{b_n }} - f'(0)} \right| + \left| {\frac{{f(a_n ) - f(0)}}{{a_n }} - f'(0)} \right|.
$$
Thus,
\begin{align*} &
\mathop {\lim }\limits_{n \to  + \infty } \left| {D_n  - f'(0)} \right| \le \mathop {\lim }\limits_{n \to  + \infty } \left| {\frac{{f(b_n ) - f(0)}}{{b_n }} - f'(0)} \right| + \mathop {\lim }\limits_{n \to  + \infty } \left| {\frac{{f(a_n ) - f(0)}}{{a_n }} - f'(0)} \right|
\\ &
 \le \left| {\mathop {\lim }\limits_{n \to  + \infty } \frac{{f(b_n ) - f(0)}}{{b_n }} - f'(0)} \right| + \left| {\mathop {\lim }\limits_{n \to  + \infty } \frac{{f(a_n ) - f(0)}}{{a_n }} - f'(0)} \right|
\\ &
 = \left| {f'(0) - f'(0)} \right| + \left| {f'(0) - f'(0)} \right| = 0.
\end{align*}
