How do I calculate f'(a) = g(a)?

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Homework Statement



Problem:
1C-2 Let f(x) = (x - a)g(x). Use the definition of the derivative to calculate that f'(a) = g(a), assuming that g is continuous

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The Attempt at a Solution



I just started a self study of calculus using Khan and MIT OpenCourse. I am bit stuck, or maybe confused, with this exercise from MIT 18.01

The definition for the derivative is
[itex]\frac{Δy}{Δx}[/itex]=[itex]\frac{f(x)-f(a)}{x-a}[/itex]

I can replace f(x) with (x-a)g(x), but I don't know what to do with f(a). In the given solution f(a) somehow becomes 0, but I can't see how.

The given solution:
Equation 1C-2: [f(x) - f(a)]/(x - a) = [(x - a)g(x) - 0]/(x - a) = g(x) → g(a) as x → a.


Any help is much appreciated.
 
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  • Printed derivation: (f(x) - f(a))/(x - a) = ((x - a)g(x) - 0)/(x - a) = g(x) → g(a) as x → a.
    1C-2-solution.png
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