- #1

deancodemo

- 20

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In order to make this proof valid, we need to find a way to determine the derivative of sinx that does not rely on [itex]\lim_{x \to 0} \frac{\sin x}{x} = 1[/itex] to be true.

If we use Euler's formula, we find that [itex]\sin x = \frac{e^{ix} - e^{-ix}}{2i}[/itex] and [itex]\cos x = \frac{e^{ix} + e^{-ix}}{2}[/itex].

This means that [itex]\frac{d}{dx} \sin x = \frac{d}{dx} \frac{e^{ix} - e^{-ix}}{2i}[/itex]. But [itex]\frac{d}{dx} \frac{e^{ix} - e^{-ix}}{2i} = \frac{e^{ix}+e^{-ix}}{2}[/itex] (using the quotient rule) which is cosx. Therefore, [itex]\frac{d}{dx}\sin x = \cos x[/itex]. This was found without the assumption of [itex]\lim_{x \to 0} \frac{\sin x}{x} = 1[/itex] at all. This means we can now use L'Hôpital's rule to find the limit without constituting circular reasoning.

Is my reasoning above correct? This proof would be invalid is Euler's formula relies on [itex]\lim_{x \to 0} \frac{\sin x}{x} = 1[/itex] but I am not sure.