Derivative of a single valued function as a limit of a function in $\mathbb{R}^2$ I am very familiar to the usual definition of derivative of a function $f(x) : \mathbb{R} \rightarrow \mathbb{R}^n$ as $\lim_{x \rightarrow a}\frac{f(x)-f(a)}{(x-a)}$. My question is when can I relax the condition of "fixing a $a$"? That is, when the following also holds:
$$\lim_{(x,y)\rightarrow(a,a)} \dfrac{f(x)-f(y)}{x-y} = f'(a) \;\text{?}$$
I am convinced that this will work most of the time but I'm having trouble proving it.
 A: We suppose only that $f'(a)$ exists and ask if that limit exists and equals $f'(a)$.
If you require that $y<a<x$, then you can show that this limit does exist and equals $f'(a)$. You proceed by writing
\begin{multline*}
\left|\frac{f(x)-f(y)}{x-y} - f'(a)\right| \le \left|\left(\frac{f(x)-f(a)}{x-a}-f'(a)\right)\left(\frac{x-a}{x-y}\right)\right| \\ + \left|\left(\frac{f(a)-f(y)}{a-y}-f'(a)\right)\left(\frac{a-y}{x-y}\right)\right|,
\end{multline*}
and then noting that $\left|\frac{x-a}{x-y}\right|$ and $\left|\frac{a-y}{x-y}\right|$ are both less than $1$.
However, if $a<y<x$, it need not work out. Take a function $f$ that is differentiable at $a$ but not $C^1$. The standard example is
$$f(x) = \begin{cases} x^2\sin(1/x), & x\ne 0 \\ 0, & x=0\end{cases}$$
with $a=0$. Take $x_n = \dfrac1{2\pi n}$ and $y_n = \dfrac1{2\pi n + \pi/2}$; then $x_n-y_n = \dfrac{\pi/2}{(2\pi n)(2\pi n+\pi/2)}$. Thus, we have
$$\frac{f(x_n)-f(y_n)}{x_n-y_n} = \frac{y_n^2}{x_n-y_n} = \frac2\pi\cdot\frac{2\pi n}{2\pi n + \pi/2}.$$
It certainly follows that $\lim\limits_{n\to\infty} \dfrac{f(x_n)-f(y_n)}{x_n-y_n} = \dfrac2\pi \ne 0 = f'(0)$.
