Equivalent form of derivative as limit? I was traditionally taught the formula for the derivative to be:
$$ \dfrac{df}{dx} = \lim_{\Delta x \to 0} \dfrac{f(x + \Delta x) - f(x)}{\Delta x}$$
Is this an equally valid form? How can I see one way or the other?
$$ \dfrac{df}{dx} \overset{?}{=} \lim_{\Delta x \to 0} \dfrac{f(x) - f(x - \Delta x)}{\Delta x}$$
 A: Let $\Delta y=-\Delta x$. Then 
$$\lim_{\Delta x \to 0} \dfrac{f(x) - f(x - \Delta x)}{\Delta x}= \lim_{\Delta y \to 0} \dfrac{f(x) - f(x + \Delta y)}{-\Delta y}$$
Now move the $-$ to the numerator.
A: You can derive $\infty$ expressions (Using Taylor series expansion of $f(x+h)$ around $x=x$) !!!!. For example
\begin{eqnarray*}
{\rm f}\left(x + h\right)
& = &
{\rm f}\left(x\right) + {\rm f}'\left(x\right)h + {1 \over 2}{\rm f}''\left(x\right)h^{2}
+
{1 \over 6}{\rm f}'''\left(x\right)h^{3} + \cdots
\\
{\rm f}\left(x + 2h\right)
& = &
{\rm f}\left(x\right) + 2{\rm f}'\left(x\right)h + 2{\rm f}''\left(x\right)h^{2}
-
{4 \over 3}{\rm f}'''\left(x\right)h^{3} + \cdots
\\
8{\rm f}\left(x + h\right) + {\rm f}\left(x + 2h\right)
& = &
9{\rm f}\left(x\right) + 10{\rm f}'\left(x\right)h + 6{\rm f}''\left(x\right)h^{2}
+
{\rm O}\left(h^{4}\right)
\end{eqnarray*}
Then
$$
{\rm f}'\left(x\right)
=
\lim_{h \to 0}
{8{\rm f}\left(x + h\right) + {\rm f}\left(x + 2h\right) - 9{\rm f}\left(x\right)
 \over
 10 h}
$$
A: This does not shed light on what's going on as much as the previous answer and comments, but it's worth observing that by taking either limit as the definition of the derivative, you can evaluate the other limit using L'Hospital's rule. 
