Limit of (Cosine (X) - 1) over X/Proof 3

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Theorem

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


Proof

\(\displaystyle \lim_{x \mathop \to 0} \ \frac {\cos x - 1} x\) \(=\) \(\displaystyle \lim_{x \mathop \to 0} \ \frac {\left({\cos x - 1}\right) \left({\cos x + 1}\right)} {x \left({\cos x + 1}\right)}\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{x \mathop \to 0} \ \frac {\cos^2 x - 1} {x \left({\cos x + 1}\right)}\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{x \mathop \to 0} \ \frac {-\sin^2 x} {x \left({\cos x + 1}\right)}\) 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
\(\displaystyle \) \(=\) \(\displaystyle \frac 0 2\)
\(\displaystyle \) \(=\) \(\displaystyle 0\)

$\blacksquare$