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

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Theorem

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


Proof 1

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 of Product
\(\ds \) \(\le\) \(\ds \frac 1 {2^{2 n - 1} \paren {2 n - 1}!}\) 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:

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

$\blacksquare$


Proof 2

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


From Cosine of Zero is One:

$\cos 0 = 1$

From Derivative of Cosine Function:

$\map {D_x} {\cos x} = -\sin x$

and by Derivative of Constant:

$\map {D_x} {-1} = 0$

So by Sum Rule for Derivatives:

$\map {D_x} {\cos x - 1} = -\sin x$

By Sine of Zero is Zero:

$\sin 0 = 0$

From Derivative of Identity Function:

$\map {D_x} x = 1$


Thus L'Hôpital's Rule applies and so:

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

$\blacksquare$


Proof 3

\(\ds \lim_{x \mathop \to 0} \frac {\cos x - 1} x\) \(=\) \(\ds \lim_{x \mathop \to 0} \frac {\paren {\cos x - 1} \paren {\cos x + 1} } {x \paren {\cos x + 1} }\)
\(\ds \) \(=\) \(\ds \lim_{x \mathop \to 0} \frac {\cos^2 x - 1} {x \paren {\cos x + 1} }\)
\(\ds \) \(=\) \(\ds \lim_{x \mathop \to 0} \frac {-\sin^2 x} {x \paren {\cos x + 1} }\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds \paren {\lim_{x \mathop \to 0} \frac {\sin x} x} \paren {\lim_{x \mathop \to 0} \frac {-\sin x} {\cos x + 1} }\) Product Rule for Limits of Real Functions
\(\ds \) \(=\) \(\ds 1 \times \paren {\lim_{x \mathop \to 0} \frac{-\sin x} {\cos x + 1} }\) Limit of $\dfrac {\sin x} x$ at Zero
\(\ds \) \(=\) \(\ds \frac {\ds \lim_{x \mathop \to 0} \paren {-\sin x} } {\ds \lim_{x \mathop \to 0} \paren {\cos x + 1} }\) Quotient Rule for Limits of Real Functions
\(\ds \) \(=\) \(\ds \frac 0 2\)
\(\ds \) \(=\) \(\ds 0\)

$\blacksquare$


Proof 4

\(\ds \frac {\cos x - 1} x\) \(=\) \(\ds \frac {\cos x - \cos 0} x\) Cosine of Zero is One
\(\ds \) \(\to\) \(\ds \valueat {\dfrac \d {\d x} \cos x} {x \mathop = 0}\) as $x \to 0$, from Definition of Derivative of Real Function at Point
\(\ds \) \(=\) \(\ds \bigvalueat {\sin x} {x \mathop = 0}\) Derivative of Cosine Function
\(\ds \) \(=\) \(\ds 0\) Sine of Zero is Zero

$\blacksquare$

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