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

Theorem

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

Proof

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

\(\ds \cos x\)

\(=\)

\(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}\)

Definition of Real Cosine Function

\(\ds \)

\(=\)

\(\ds \paren {-1}^0 \cdot \frac {x^{2 \cdot 0} } {\paren {2 \cdot 0}!} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}\)

\(\ds \)

\(=\)

\(\ds 1 + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}\)

Definition of Zero Factorial and Definition of Zeroth Power

\(\ds \lim_{x \mathop \to 0} \frac {\cos x - 1} x\)

\(=\)

\(\ds \lim_{x \mathop \to 0} \frac {1 + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} - 1} x\)

\(\ds \)

\(=\)

\(\ds \lim_{x \mathop \to 0} \frac {\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} } x\)

\(\ds \)

\(=\)

\(\ds \lim_{x \mathop \to 0} \frac {\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n - 1} } {\paren {2 n - 1}!} } 1\)

Power Series is Differentiable on Interval of Convergence and L'Hôpital's Rule

\(\ds \)

\(=\)

\(\ds \lim_{x \mathop \to 0} \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n - 1} } {\paren {2 n - 1}!}\)

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

\(\ds \map {f_n} x\)

\(\le\)

\(\ds \size {\paren {-1}^n \frac {x^{2 n - 1} } {\paren {2 n - 1}!} }\)

\(\ds \; = \frac { {\size x}^{2 n - 1} } {\paren {2 n - 1}!}\)

Absolute Value Function is Completely Multiplicative

\(\ds \)

\(\le\)

\(\ds \frac 1 {2^{2 n - 1} \paren {2 n - 1}!}\)

Power Function is Strictly Increasing over Positive Reals

\(\ds \)

\(\le\)

\(\ds \frac 1 {2^{2 n - 1} }\)

because the factorial is strictly increasing

\(\ds \)

\(\le\)

\(\ds \frac 1 {2^n}\)

because $n \ge 1 \iff 2 n - 1 \ge n$

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:

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

$\blacksquare$