How do you find the limit of $\frac{1 - cos x}{x}$ $(1-cosx) / x$ as x approaches 0?

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Answer 1 · 2016-10-03T15:41:00

$0$ $0$

$1 - cos x = 2 {sin}^{2} (\frac{x}{2})$ $1-cosx=2sin^2(x/2)$ so

$\frac{1 - cos x}{x} = (\frac{x}{4}) {(\frac{sin (\frac{x}{2})}{\frac{x}{2}})}^{2}$ $(1-cos x)/x=(x/4) (sin(x/2)/(x/2))^2$ then

$lim_(x->0)(1-cos x)/x equiv lim_(x->0)(x/4) (sin(x/2)/(x/2))^2 = 0 cdot 1 = 0$

Answer 2 · 2016-10-03T16:25:09

Multiply by $(1+cosx)/(1+cosx)$ to get

$(1-cos^2x)/(x(1+cosx)) = (sin^2x)/(x(1+cosx))$

$= sinx * sinx/x * 1/(1+cosx)$

Taking the limit as $xrarr0$ gives

$(0)(1)(1/2) = 0$

How do you find the limit of (1-cosx) / x1−cosxx(1-cosx) / x as x approaches 0?