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

Question

3610 views around the world

You can reuse this answer
Creative Commons License

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.