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 $\displaystyle f_n \left({x}\right) = \left({-1}\right)^n \frac {x^{2n-1}}{\left({2n-1}\right)!}$

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
 * $\displaystyle \sum_{n \mathop = 1}^\infty \frac {1}{2^n} = 2 \lt \infty$

And so by the Weierstrass M-Test, $\sum_{n=1}^\infty f_n \left({x}\right)$ 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
 * $\displaystyle \lim_{x \to 0} f \left({x}\right) = f \left({0}\right) = \sum_{n \mathop = 1}^\infty \left({-1}\right) \frac {0^{2n-1}}{\left({2n-1}\right)!} = 0$