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

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Answer 1 · 2017-11-12T02:23:05

$lim_(x rarr 0) (1- cosx)/(x sinx) = 1/2$

First of all, since as $x rarr 0$ , $sinx rarr 0$ also, we can rewrite the denominator as $x^2$ .

Hence we need to find: $lim_(x rarr 0) (1- cosx)/(x^2)$

Since this still results in an indeterminate $0/0$ , we apply L'Hopital's Rule.

$(d/dx(1-cos x)) / (d/dx(x^2)) = sinx/(2x)$

If we substitute 'approaching zero' as a less formal $1/oo$ , we arrive at the expression: $(1/oo)/(2/oo)$ .

After cancelling out the infinities, this leaves $1/2$ .

Or simply let $sinx =x$ again, which gives:

$sinx/(2x) = x/(2x) = 1/2$ .

Answer 2 · 2017-11-12T02:42:11

Use $lim_(xrarr0)sinx/x=1$

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

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

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

$lim_(x rarr 0) (1- cosx)/(x sinx) = lim_(x rarr 0)(sinx/x * sinx/sinx * 1/(1+cosx))$

$= (1)(1)(1/(1+1))=1/2$

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