Just making sure I understood:
$$\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}=f'(x_0)$$
At a first glance I didn't understand why the above is true. It's because (in the case above) we can say that $x=x_0+\Delta x$; right?
Just making sure I understood:
$$\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}=f'(x_0)$$
At a first glance I didn't understand why the above is true. It's because (in the case above) we can say that $x=x_0+\Delta x$; right?
This is what I use as the definition of the derivative. However, if you define the derivative as $$f'(x_0):=\lim\limits_{\Delta x\to 0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}$$ then your statement is correct.
I've seen the definition written as $$ \frac{dy}{dx} = \lim_{\Delta x\to0} \frac{\Delta y}{\Delta x}, $$ which is also equivalent to the forms discussed above.
If the definition you've seen is $$ f'(x_0) = \lim_{\Delta x\to0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x_0} $$ (or you might call it $h$ instead of $\Delta x$) then you can say $$ \begin{align} x & = x_0+\Delta x, \\[12pt] \text{so }\Delta x & = x - x_0, \\[12pt] \text{and as }\Delta x & \to0,\text{ then }x \to x_0, \\ \end{align} $$ and one form becomes the other.