How do you evaluate the limit $(3(1-cosx))/x$ as x approaches $0$ ?

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Answer 1 · 2016-09-02T12:25:34

$0$

$lim_(x to 0) (3(1-cosx))/x$

we're going to steer this in the direction of well known limit $lim_(z to 0) (sin z)/z = 1$

using the double angle identity: $cos 2A = 1 - sin^2 A$

$= lim_(x to 0) (3(1-(1 - 2 sin^2 (x/2))))/x$

$= lim_(x to 0) (6 sin^2 (x/2))/x$

pattern matching numerator and denominator, and taking the constant term outside the limit
$=12 lim_(x to 0) ( sin^2 (x/2))/(x/2)$

$=12 lim_(x to 0) ( sin (x/2))/(x/2) * sin (x/2)$

these limits of these individual terms exist and the limit of the product is the product of the limits

$=12 lim_(x to 0) ( sin (x/2))/(x/2) * lim_(x to 0) sin (x/2)$

$=12 * 1* 0 = 0$

You cannot use L'Hopital's Rule for this.

How do you evaluate the limit (3(1-cosx))/x(3(1-cosx))/x as x approaches 00?