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**Deveno** the continuity of g is also why the term $\displaystyle \lim_{\Delta x \to 0} \frac{\Delta f \Delta g}{\Delta x}$ vanishes, since:

$\displaystyle \lim_{\Delta x \to 0} \frac{\Delta f\Delta g}{\Delta x} = \left(\lim_{\Delta x \to 0} \frac{\Delta f}{\Delta x}\right)(\lim_{\Delta x \to 0} \Delta g)$

$\displaystyle = f'(x_0)(\lim_{\Delta x \to 0} \Delta g)$

and recalling that:

$\displaystyle \Delta g = g(x_0 + \Delta x) - g(x_0)$

$\displaystyle \lim_{\Delta x \to 0} \Delta g = (\lim_{\Delta x \to 0} g(x_0 + \Delta x)) - (\lim_{\Delta x \to 0} g(x_0))$

and *because g is continuous*

$\displaystyle = g(x_0) - g(x_0) = 0$

(the continuity of g is essential, if g suddenly "jumps" at x_{0}, we cannot say that g(x_{0}+Δx) is anywhere near g(x_{0})).