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:
 * $f_n \left({x}\right) = \left({-1}\right)^n \dfrac {x^{2n - 1} } {\left({2 n - 1}\right)!}$

Then for every $n \in \mathbb N_{\gt 0}$, and for all $x \in \left[{\frac 1 2 \,.\,.\, \frac 1 2}\right]$:

But from the Geometric Series we have
 * $\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {2^n} = 2 < \infty$

By the Weierstrass M-Test, $\displaystyle \sum_{n \mathop = 1}^\infty f_n \left({x}\right)$ converges uniformly to some function $f$ on $\left[{\frac 1 2 \,.\,.\, \frac 1 2}\right]$.

But from Polynomial is Continuous, and the Uniform Limit Theorem $f$ is continuous on $\left[{\frac 1 2 \,.\,.\, \frac 1 2}\right]$.

So:
 * $\displaystyle \lim_{x \mathop \to 0} f \left({x}\right) = f \left({0}\right) = \sum_{n \mathop = 1}^\infty \left({-1}\right) \frac {0^{2 n - 1} } {\left({2 n - 1}\right)!} = 0$