Question 475229
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ \frac{1}{x}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x\,+\,h)\ =\ \frac{1}{x\,+\,h}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x\,+\,h)\ -\ f(x)\ =\ \frac{1}{x\,+\,h}\ -\ \frac{1}{x}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x\,+\,h)\ -\ f(x)\ =\ \frac{x\ -\ (x\ +\ h)}{x^2\ +\ xh}\ =\ \frac{-h}{x^2\ +\ xh}]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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