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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?

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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.

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    I actually did try to accept yours, somewhy the system told me to wait (at the time, now I can and did). Anyway this question/answer now seems too obvious to even ask it (apparently I was very tired yesterday).2012-09-10
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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.