User:Thpigdog/Eulers formula proof based on De Moivres formula

Eulers formula, for real x, may be obtained from De Moivres formula, for integer n,
 * $ (\cos(\theta) + i \sin(\theta))^n = \cos(n \theta) + i \sin(n \theta) $

Let $\theta = \frac{x}{n}$, and take the limit as n tends to infinity;


 * $ \displaystyle \lim_{n \to \infty}(\cos(\frac{x}{n}) + i \sin(\frac{x}{n}))^n = \cos(\frac{n x}{n}) + i \sin(\frac{n x}{n}) $

Using Taylor's Theorem,
 * $ \cos(\frac{x}{n}) = 1 - \frac{0}{n} - \frac 1 {\left(2\right)!} \left({\frac{x}{n}}\right)^2 \cos \left({\eta_1}\right) $
 * $ \sin(\frac{x}{n}) = 0 + \frac{x}{n} - \frac 1 {\left(2\right)!} \left({\frac{x}{n}}\right)^2 \sin \left({\eta_2}\right) $

gives,
 * $ \displaystyle \lim_{n \to \infty} (\cos(\frac{x}{n}) + i \sin(\frac{x}{n}))^n = \lim_{n \to \infty} (1 + \frac{i x}{n} - \frac 1 {\left(2\right)!} \left({\frac{x}{n}}\right)^2 (\cos \left({\eta_1}\right) + i \sin \left({\eta_2}\right)))^n$

Using User:Thpigdog/Limit power identity,
 * $ \displaystyle \lim_{n \to \infty}(1 + \frac{a}{n} + \frac{b}{n^2})^n = \lim_{n \to \infty}(1 + \frac{a}{n})^n $

gives
 * $ \displaystyle \lim_{n \to \infty} (1 + \frac{i x}{n} - \frac 1 {\left(2\right)!} \left({\frac{x}{n}}\right)^2 (\cos \left({\eta_1}\right) + i \sin \left({\eta_2}\right)))^n = \lim_{n \to \infty} (1 + \frac{i x}{n})^n$

so,
 * $ \displaystyle \lim_{n \to \infty}((\cos(\frac{x}{n}) + i \sin(\frac{x}{n}))^n = \lim_{n \to \infty}((1 + \frac{i x}{n}) = \cos(x) + i \sin(x)$

From the Equivalence of Definitions of Exponential Function,
 * $ \displaystyle \lim_{n \to \infty}(1+\frac{z}{n})^n = \sum_{k=0}^\infty \frac{z^k}{k!} = e^z $

so where $$z = i x$$
 * $ \displaystyle \lim_{n \to \infty}(1 + \frac{i x}{n})^n = e^{i x} = \cos(x) + i \sin(x)$