So here is the question:
Suppose that $f:(-1,1)\to\mathbb{R}$ is differentiable at $0$ and that $(x_n)$ and $(y_n)$ are sequences in $(-1,1)$ such that $x_n\to 0, y_n \to 0$, and $x_n < y_n$ for all $n$. Let $$D_n = \frac{f(y_n)-f(x_n)}{y_n - x_n}$$
Show that if $f'$ is defined and continuous in $(-1,1)$ then $D_n \to f'(0)$
I've thought to prove this by holding $x_n$ fixed and take $y_n\to0$, then take $x_n\to0$.
But I then have a concern that the limit by taking $x_n\to0,y_n\to0$ separately may not be equal to the limit by taking $x_n\to0,y_n\to0$ at the same time.
Is there another approach for this?
Or is there any evidence to show that the limits by taking $x_n,y_n\to0$ separately or at the same time are the same?
Hint:
By mean value theorem, for every $n$ there exists a $c_n\in(x_n,y_n)$ with $$ D_n = \frac{f(y_n) - f(x_n)}{y_n - x_n} = f'(c_n).$$ Now, as $y_n - x_n \to 0$ it follows $c_n \to 0$ and $D_n\to...?$