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

Theorem

$\displaystyle \lim_{x \to 0} \frac {\cos \left({x}\right) - 1} {x} = 0$

Proof 1

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \cos x\)	\(=\)	\(\displaystyle \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n} }{\left({2n}\right)!}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from the definition of the cosine function
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle (-1)^0 \cdot \frac{x^{2\cdot0} }{(2\cdot0)!}+\sum_{n=1}^\infty \left({-1}\right)^n \frac {x^{2n} }{\left({2n}\right)!}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1 + \sum_{n=1}^\infty \left({-1}\right)^n \frac {x^{2n} }{\left({2n}\right)!}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from the definition of $0!$ and the definition of $a^0$

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

$\blacksquare$

Proof 2

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

We have that:

• From Cosine of Zero is One: $\cos 0 = 1$
• From Derivative of Cosine Function: $D_x \left({\cos x}\right) = - \sin x$ and by Derivative of Constant: $D_x \left({-1}\right) = 0$. So by Sum Rule for Derivatives $D_x \left({\cos x - 1}\right) = - \sin x$
• By Sine of Zero is Zero, $\sin 0 = 0$
• From Derivative of Identity Function: $D_x \left({x}\right) = 1$.

Thus L'Hôpital's Rule applies and so $\displaystyle \lim_{x \to 0} \frac {\cos x - 1} x = \lim_{x \to 0} \frac {-\sin x} 1 = \frac {-0} 1 = 0$.

$\blacksquare$

Proof 3

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

$\blacksquare$