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

Eulers formula, for real $x$, may be obtained from De Moivres formula, for integer $n$:

$\paren {\map \cos \theta + i \map \sin \theta}^n = \map \cos {n \theta} + i \map \sin {n \theta}$

Let $\theta = \dfrac x n$, and take the limit as n tends to infinity:

$\ds \lim_{n \mathop \to \infty} \paren {\map \cos {\dfrac x n} + i \map \sin {\dfrac x n} }^n = \map \cos {\dfrac {n x} n} + i \map \sin {\frac {n x} n}$

Using Taylor's Theorem,

$\map \cos {\dfrac x n} = 1 - \dfrac 0 n - \dfrac 1 {\paren 2!} \paren {\dfrac x n}^2 \map \cos {\eta_1}$

$\map \sin {\dfrac x n} = 0 + \dfrac x n - \dfrac 1 {\paren 2!} \paren {\dfrac x n}^2 \map \sin {\eta_2}$

gives:

$\ds \lim_{n \mathop \to \infty} \paren {\map \cos {\dfrac x n} + i \map \sin {\dfrac x n} }^n = \lim_{n \mathop \to \infty} \paren {1 + \dfrac {i x} n - \dfrac 1 {\paren 2!} \paren {\dfrac x n}^2 \paren {\map \cos {\eta_1} + i \map \sin {\eta_2} } }^n$

Using User:Thpigdog/Limit power identity,

$\ds \lim_{n \mathop \to \infty} \paren {1 + \dfrac a n + \dfrac b {n^2} }^n = \lim_{n \mathop \to \infty} \paren {1 + \dfrac a n)}^n$

gives:

$\ds \lim_{n \mathop \to \infty} \paren {1 + \dfrac {i x} n - \dfrac 1 {\paren 2!} \paren {\dfrac x n}^2 \paren {\map \cos {\eta_1} + i \map \sin {\eta_2} } }^n = \lim_{n \mathop \to \infty} \paren {1 + \dfrac {i x} n}^n$

so:

$\ds \lim_{n \mathop \to \infty} \paren {\map \cos {\dfrac x n} + i \map \sin {\dfrac x n} }^n = \lim_{n \mathop \to \infty} \paren {1 + \dfrac {i x} n}^n = \map \cos x + i \map \sin x$

From the Equivalence of Definitions of Exponential Function:

$\ds \lim_{n \mathop \to \infty} \paren {1 + \dfrac z n}^n = \sum_{k \mathop = 0}^\infty \dfrac {z^k} {k!} = e^z$

so where $z = i x$:

$\ds \lim_{n \mathop \to \infty} \paren {1 + \frac {i x} n}^n = e^{i x} = \map \cos x + i \map \sin x$