limit of ${\displaystyle \frac{1-cosx}{x}}$ as $x$ approaches 0

Corollary 1.

$$\underset{x\to 0}{lim}\frac{1-\cos x}{x}=0$$

Proof.

$\underset{x\to 0}{lim}{\displaystyle \frac{1-\cos x}{x}}$	$=\underset{x\to 0}{lim}{\displaystyle \frac{(1-\cos x)(1+\cos x)}{x(1+\cos x)}}$
	$=\underset{x\to 0}{lim}{\displaystyle \frac{1-\cos x+\cos x-{\cos }^{2}x}{x(1+\cos x)}}$
	$=\underset{x\to 0}{lim}{\displaystyle \frac{1-{\cos }^{2}x}{x(1+\cos x)}}$
	$=\underset{x\to 0}{lim}{\displaystyle \frac{{\sin }^{2}x}{x(1+\cos x)}}$
	$=\underset{x\to 0}{lim}{\displaystyle \frac{\sin x}{x}}\cdot {\displaystyle \frac{\sin x}{1+\cos x}}$
	$=\left(\underset{x\to 0}{lim}{\displaystyle \frac{\sin x}{x}}\right)\left(\underset{x\to 0}{lim}{\displaystyle \frac{\sin x}{1+\cos x}}\right)$ by this entry (http://planetmath.org/LimitRulesOfFunctions)
	$=1\cdot \underset{x\to 0}{lim}{\displaystyle \frac{\sin x}{1+\cos x}}$ by this entry (http://planetmath.org/LimitOfDisplaystyleFracsinXxAsXApproaches0)
	$={\displaystyle \frac{\sin 0}{1+\cos 0}}$ by this entry (http://planetmath.org/LimitRulesOfFunctions) and the fact that $\sin$ and $\cos$ are continuous
	$={\displaystyle \frac{0}{1+1}}$
	$=0$ ∎

This corollary has an obvious corollary to it:

Corollary 2.

$$\underset{x\to 0}{lim}\frac{\cos x-1}{x}=0$$

Title	limit of ${\displaystyle \frac{1-cosx}{x}}$ as $x$ approaches 0
Canonical name	LimitOfdisplaystylefrac1cosXxAsXApproaches0
Date of creation	2013-03-22 16:58:49
Last modified on	2013-03-22 16:58:49
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	7
Author	Wkbj79 (1863)
Entry type	Corollary
Classification	msc 26A06
Classification	msc 26A03
Related topic	DerivativesOfSinXAndCosX