How do you find the limit of #(x-sin abs[x])/(3x^3) # as x approaches #0#?

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Answer 1 · 2016-07-09T17:11:39

#lim_{x to 0^-} (x-sin abs[x])/(3x^3) = oo#

#lim_{x to 0^+} (x-sin abs[x])/(3x^3) = 1/18#

#lim_{x to 0} (x-sin abs[x])/(3x^3)#

that #abs x # thing is a real pain so we can consider the different forms.

for x < 0, the limit becomes

#lim_{x to 0^-} (x-sin (-x))/(3x^3)#

#= lim_{x to 0^-} (x +sin x)/(3x^3) = 0/0# indeterminate so we use LHopital

#= lim_{x to 0^-} (1 +cos x)/(9x^2) #

#cos x# is continuous at x = 0 so the limit is #2/0 = oo#

for x > 0, the limit becomes

#lim_{x to 0^+} (x-sin x)/(3x^3) = 0/0#

indeterminate so again we use LHopital

#= lim_{x to 0^+} (1-cos x)/(9x^2) = 0/0#

so we go again

#=lim_{x to 0^+} (sin x)/(18x) = 1/18 lim_{x to 0^+} (sin x)/(x) = 1/18#

as #lim_{x to 0} (sin x)/(x) = 1#