# limit of $\displaystyle\frac{1-\mathop{cos}\nolimits x}{x}$ as $x$ approaches 0

###### Corollary 1.
 $\lim_{x\to 0}\frac{1-\cos x}{x}=0$
###### Proof.
$\displaystyle\lim_{x\to 0}\frac{1-\cos x}{x}$ $=\displaystyle\lim_{x\to 0}\frac{(1-\cos x)(1+\cos x)}{x(1+\cos x)}$ $=\displaystyle\lim_{x\to 0}\frac{1-\cos x+\cos x-\cos^{2}x}{x(1+\cos x)}$ $=\displaystyle\lim_{x\to 0}\frac{1-\cos^{2}x}{x(1+\cos x)}$ $=\displaystyle\lim_{x\to 0}\frac{\sin^{2}x}{x(1+\cos x)}$ $=\displaystyle\lim_{x\to 0}\frac{\sin x}{x}\cdot\frac{\sin x}{1+\cos x}$ $=\displaystyle\left(\lim_{x\to 0}\frac{\sin x}{x}\right)\left(\lim_{x\to 0}% \frac{\sin x}{1+\cos x}\right)$ by this entry (http://planetmath.org/LimitRulesOfFunctions) $=\displaystyle 1\cdot\lim_{x\to 0}\frac{\sin x}{1+\cos x}$ 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.
 $\lim_{x\to 0}\frac{\cos x-1}{x}=0$
Title limit of $\displaystyle\frac{1-\mathop{cos}\nolimits x}{x}$ as $x$ approaches 0 LimitOfdisplaystylefrac1cosXxAsXApproaches0 2013-03-22 16:58:49 2013-03-22 16:58:49 Wkbj79 (1863) Wkbj79 (1863) 7 Wkbj79 (1863) Corollary msc 26A06 msc 26A03 DerivativesOfSinXAndCosX