# Limit of Sine of X over X/Proof 1

## Theorem

$\displaystyle \lim_{x \mathop \to 0} \frac {\sin x} x = 1$

## Proof

 $\displaystyle \sin x$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2n+1} }{\left({2n+1}\right)!}$ Definition of the Sine Function $\displaystyle$ $=$ $\displaystyle \left({-1}\right)^0 \frac{x^{2 \cdot 0 + 1} } { \left({2 \cdot 0 + 1}\right)!} + \sum_{n \mathop = 1}^\infty \left({-1}\right)^n \frac {x^{2n+1} } {\left({2n+1}\right)!}$ $\displaystyle$ $=$ $\displaystyle x + \sum_{n \mathop = 1}^\infty \left({-1}\right)^n \frac {x^{2n+1} }{\left({2n+1}\right)!}$

 $\displaystyle \lim_{x \mathop \to 0}\frac{\sin x} x$ $=$ $\displaystyle \lim_{x \mathop \to 0} \frac{x + \sum_{n \mathop = 1}^\infty \left({-1}\right)^n \frac {x^{2n+1} }{\left({2n+1}\right)!} } x$ $\displaystyle$ $=$ $\displaystyle \lim_{x \mathop \to 0} \frac x x + \lim_{x \mathop \to 0} \frac{\sum_{n \mathop = 1}^\infty \left({-1}\right)^n \frac {x^{2n + 1} }{\left({2n+1}\right)!} } x$ $\displaystyle$ $=$ $\displaystyle 1 + \lim_{x \mathop \to 0} \frac{\sum_{n \mathop = 1}^\infty \left({-1}\right)^n \frac {x^{2n} } {\left({2n}\right)!} } 1$ Power Series is Differentiable on Interval of Convergence and L'Hôpital's Rule $\displaystyle$ $=$ $\displaystyle 1 + \lim_{x \mathop \to 0} \sum_{n \mathop = 1}^\infty \left({-1}\right)^n \frac {x^{2n} }{\left({2n}\right)!}$ $\displaystyle$ $=$ $\displaystyle 1 + \sum_{n \mathop = 1}^\infty \left({-1}\right)^n \frac {0^{2n} }{\left({2n}\right)!}$ Polynomial is Continuous $\displaystyle$ $=$ $\displaystyle 1$

$\blacksquare$