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

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Answer 1 · 2016-10-03T15:41:00

#0#

#1-cosx=2sin^2(x/2)# so

#(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#