Limit of (Cosine (X) - 1) over X at Zero/Proof 1

Proof
This proof works directly from the definition of the cosine function:

Now let:
 * $\map {f_n} x = \paren {-1}^n \dfrac {x^{2 n - 1} } {\paren {2 n - 1}!}$

Then for every $n \in \N_{> 0}$, and for all $x \in \closedint {\dfrac 1 2} {\dfrac 1 2}$:

But from Sum of Infinite Geometric Sequence:
 * $\ds \sum_{n \mathop = 1}^\infty \frac 1 {2^n} = 2 < \infty$

By the Weierstrass M-Test, $\ds \sum_{n \mathop = 1}^\infty \map {f_n} x$ converges uniformly to some function $f$ on $\closedint {\dfrac 1 2} {\dfrac 1 2}$.

But from Real Polynomial Function is Continuous, and the Uniform Limit Theorem $f$ is continuous on $\closedint {\dfrac 1 2} {\dfrac 1 2}$.

So:
 * $\ds \lim_{x \mathop \to 0} \map f x = \map f 0 = \sum_{n \mathop = 1}^\infty \paren {-1} \frac {0^{2 n - 1} } {\paren {2 n - 1}!} = 0$