# Limit of (Cosine (X) - 1) over X

## Theorem

$\displaystyle \lim_{x \mathop \to 0} \frac {\map \cos x - 1} x = 0$

## Proof 1

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

 $\displaystyle \cos x$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}$ Definition of Real Cosine Function $\displaystyle$ $=$ $\displaystyle \paren {-1}^0 \cdot \frac {x^{2 \cdot 0} } {\paren {2 \cdot 0}!} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}$ $\displaystyle$ $=$ $\displaystyle 1 + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}$ Definition of Zero Factorial and Definition of Zeroth Power

 $\displaystyle \lim_{x \mathop \to 0} \frac {\map \cos x - 1} x$ $=$ $\displaystyle \lim_{x \mathop \to 0} \frac {1 + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} - 1} x$ $\displaystyle$ $=$ $\displaystyle \lim_{x \mathop \to 0} \frac {\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} } x$ $\displaystyle$ $=$ $\displaystyle \lim_{x \mathop \to 0} \frac {\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n - 1} } {\paren {2 n - 1}!} } 1$ Power Series is Differentiable on Interval of Convergence and L'Hôpital's Rule $\displaystyle$ $=$ $\displaystyle \lim_{x \mathop \to 0} \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n - 1} } {\paren {2 n - 1}!}$

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 $x \in \closedint {\dfrac 1 2} {\dfrac 1 2}$:

 $\displaystyle \map {f_n} x$ $\le$ $\displaystyle \size {\paren {-1}^n \frac {x^{2 n - 1} } {\paren {2 n - 1}!} }$ $\displaystyle \; = \frac { {\size x}^{2 n - 1} } {\paren {2 n - 1}!}$ Absolute Value Function is Completely Multiplicative $\displaystyle$ $\le$ $\displaystyle \frac 1 {2^{2 n - 1} \paren {2 n - 1}!}$ Power Function is Strictly Increasing over Positive Reals $\displaystyle$ $\le$ $\displaystyle \frac 1 {2^{2 n - 1} }$ because the factorial is strictly increasing $\displaystyle$ $\le$ $\displaystyle \frac 1 {2^n}$ because $n \ge 1 \iff 2 n - 1 \ge n$

But from Sum of Infinite Geometric Sequence:

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

By the Weierstrass M-Test, $\displaystyle \sum_{n \mathop = 1}^\infty \map {f_n} x$ 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:

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

$\blacksquare$

## Proof 2

This proof assumes the truth of the Derivative of Cosine Function:

$\cos 0 = 1$
$D_x \left({\cos x}\right) = - \sin x$

and by Derivative of Constant:

$D_x \left({-1}\right) = 0$

So by Sum Rule for Derivatives:

$D_x \left({\cos x - 1}\right) = - \sin x$
$\sin 0 = 0$
$D_x \left({x}\right) = 1$

Thus L'Hôpital's Rule applies and so:

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

$\blacksquare$

## Proof 3

 $\displaystyle \lim_{x \mathop \to 0} \ \frac {\cos x - 1} x$ $=$ $\displaystyle \lim_{x \mathop \to 0} \ \frac {\left({\cos x - 1}\right) \left({\cos x + 1}\right)} {x \left({\cos x + 1}\right)}$ $\displaystyle$ $=$ $\displaystyle \lim_{x \mathop \to 0} \ \frac {\cos^2 x - 1} {x \left({\cos x + 1}\right)}$ $\displaystyle$ $=$ $\displaystyle \lim_{x \mathop \to 0} \ \frac {-\sin^2 x} {x \left({\cos x + 1}\right)}$ Sum of Squares of Sine and Cosine $\displaystyle$ $=$ $\displaystyle \left({\lim_{x \mathop \to 0} \ \frac {\sin x} x}\right) \left({\lim_{x \mathop \to 0} \ \frac {-\sin x} {\cos x + 1} }\right)$ Product Rule for Limits of Functions $\displaystyle$ $=$ $\displaystyle 1 \times \left({\lim_{x \mathop \to 0} \ \frac{-\sin x} {\cos x + 1} }\right)$ Limit of Sine of X over X $\displaystyle$ $=$ $\displaystyle \frac {\lim_{x \mathop \to 0} \left({- \sin x}\right)} {\lim_{x \mathop \to 0} \left({\cos x + 1}\right)}$ Quotient Rule for Limits of Functions $\displaystyle$ $=$ $\displaystyle \frac 0 2$ $\displaystyle$ $=$ $\displaystyle 0$

$\blacksquare$

## Proof 4

 $\displaystyle \frac {\cos x - 1} x$ $=$ $\displaystyle \frac {\cos x - \cos 0} x$ Cosine of Zero is One $\displaystyle$ $\to$ $\displaystyle \left.{\dfrac {\mathrm d} {\mathrm dx} \cos x}\right \vert_{x \mathop = 0}$ as $x \to 0$, from definition of derivative at a point $\displaystyle$ $=$ $\displaystyle \sin x \vert_{x \mathop = 0}$ Derivative of Cosine Function $\displaystyle$ $=$ $\displaystyle 0$ Sine of Zero is Zero

$\blacksquare$