Double-Sequential Characterization of Derivatives (Abbott 5.4.7) I've been trying to work through Abbott's Understanding Analysis (2nd edition) and got stuck trying to prove a lemma required for the construction of a continuous, nowhere-differentiable function. The problem statement is as follows:

Let $f$ be defined on an open interval $J$ and assume that $f$ is
  differentiable at some $a\in J$. If $(a_n)$ and $(b_n)$ are sequences
  satisfying $a_n<a<b_n$ and $\lim a_n = a = \lim b_n$, then $$f'(a) =
 \lim_{n\rightarrow\infty}\frac{f(b_n)-f(a_n)}{b_n-a_n}$$

From the sequential characterization of limits and the definition of the derivative, we know that $$\lim_{n\rightarrow\infty}\frac{f(a)-f(a_n)}{a-a_n} = f'(a) = \lim_{n\rightarrow\infty}\frac{f(b_n)-f(a)}{b_n-a}$$
This made me think that some sort of triangle inequality argument might work here. It's always true that $$\bigg|\frac{f(b_n)-f(a_n)}{b_n-a_n}-f'(a)\bigg|\leq \bigg|\frac{f(b_n)-f(a)}{b_n-a_n}-f'(a)/2\bigg| + \bigg|\frac{f(a)-f(a_n)}{a-a_n}-f'(a)/2\bigg|$$
It seems like it should be possible to show that each of these pieces can be made arbitrarily small, but I haven't been able to find a good comparison between $\frac{f(b_n)-f(a)}{b_n-a}$ and $\frac{f(b_n)-f(a)}{b_n-a_n}$ to complete the argument. Am I missing a simple step here, or is this the wrong approach for a proof? Any advice would be greatly appreciated. Thanks!
 A: Try this: observe
$$ \frac{f(b_n) - f(a_n)}{b_n - a_n} = \frac{f(a) - f(a_n)}{a-a_n} \cdot \frac{a-a_n}{b_n - a_n} + \frac{f(b_n) - f(a)}{b_n - a}\cdot \frac{b_n - a}{b_n - a_n}$$
and
$$ f'(a) = f'(a) \frac{a-a_n}{b_n - a_n} + f'(a) \frac{b_n - a}{b_n - a_n}.$$
Let 
$$I_n = \frac{f(b_n) - f(a_n)}{b_n - a_n} - f'(a).$$ Then
$$ I_n = \left( \frac{f(a) - f(a_n)}{a-a_n} - f'(a) \right) \cdot \frac{a-a_n}{b_n - a_n} + \left( \frac{f(b_n) - f(a)}{b_n - a} - f'(a) \right)\cdot \frac{b_n - a}{b_n - a_n}$$
so that
$$|I_n| \le \left| \frac{f(a) - f(a_n)}{a-a_n} - f'(a) \right| + \left| \frac{f(b_n) - f(a)}{b_n - a} - f'(a) \right|.$$
The right-hand side tends to $0$ as $n \to \infty$.
A: Here is one way:
First a little result: Suppose $x_n,y_n \to x$ and suppose $\lambda_n$ is bounded by $B$ (doesn't necessarily converge).
Then $\lambda_n x_n + (1-\lambda_n) y_n \to x$.
To see this, consider
$|\lambda_n x_n + (1-\lambda_n) y_n -x| = |\lambda_n (x_n-x) + (1-\lambda_n) (y_n -x)| \le (2B+1)(|x_n-x| + |y_n -x|)$.
Now write
${f(b_n)-f(a_n) \over b_n -a_n} = {f(b_n)-f(a) \over b_n -a} {b_n -a \over b_n -a_n} + {f(a)-f(a_n) \over a -a_n} {a-a_n \over b_n -a_n}$.
