A Sequence of Functions Converging to the Derivative at a Point I'm reading Neal Carothers' Real Analysis and while in the process of constructing an everywhere continuous but nowhere differentiable function, he claims that $$\dfrac{f(v_n)-f(u_n)}{(v_n-u_n)} \to f'(x)$$ as $n\to \infty$ if $f$ is continuous over the entire real line and is differentiable at $x$, and if $u_n\le x \le v_n$, $u_n<v_n$, and $v_n-u_n\to 0$ as $n\to \infty$. I'm trying to show that this is true by using the following definition of the derivative: $$f'(x)=\lim_{t \mathop \to x}\dfrac{f(t)-f(x)}{t-x}$$ but I can't seem to make the math come out. I thought about using the mean value theorem but we're assuming differentiability at a point, not an interval. 
(Aside: $f$ is assumed to be uniformly continuous here, although I don't think we need that fact.)
Any help at all would be much appreciated!
 A: Note 
$$\frac{f(v_n) - f(u_n)}{v_n - u_n}  = \frac{f(v_n) - f(x)}{v_n - u_n} - \frac{f(u_n) - f(x)}{v_n - u_n} $$
$$=\frac{f(v_n) - f(x)}{v_n - x} \frac{v_n-x}{v_n - u_n}  + \frac{f(u_n) - f(x)}{u_n - x} \frac{x-u_n}{v_n - u_n}$$
Now for all $\epsilon >0$, there is $N$ so that 
$$\bigg|\frac{f(v_n) - f(x)}{v_n - x}- f'(x) \bigg|, \bigg| \frac{f(v_n) - f(x)}{v_n - x} - f'(x)\bigg| < \epsilon$$ 
Whenever $n\ge N$. Thus 
$$\bigg| \frac{f(v_n) - f(u_n)}{v_n - u_n} - f'(x)\bigg| = \bigg|\frac{f(v_n) - f(x)}{v_n - x} \frac{v_n-x}{v_n - u_n}  + \frac{f(u_n) - f(x)}{u_n - x} \frac{x-u_n}{v_n - u_n} - f'(x)\bigg|$$
$$ = \bigg|\bigg(\frac{f(v_n) - f(x)}{v_n - x} -f'(x)\bigg)\frac{v_n-x}{v_n - u_n}  + \bigg(\frac{f(u_n) - f(x)}{u_n - x}-f'(x)\bigg) \frac{x-u_n}{v_n - u_n} \bigg|$$
(That's because 
$$\frac{v_n - x}{v_n - u_n} + \frac{x-u_n}{v_n - u_n} = 1)$$
Thus 
$$\bigg| \frac{f(v_n) - f(u_n)}{v_n - u_n} - f'(x)\bigg| \leq \epsilon \frac{v_n - x}{v_n - u_n} + \epsilon \frac{x-u_n}{v_n - u_n} = \epsilon$$
Whenever $n \geq N$. 
(Strictly speaking we have to consider separately the case that $u_n = x$ and $v_n = x$ for some $n$, but that is easier. )
