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

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Answer 1 · 2016-09-02T22:27:38

#=3#

we use well known limit #lim_(u to 0) (sin u)/(u) = 1#

and here we have

#lim_(x to 0) sin(3x)/x#

#= lim_(x to 0) 3 sin(3x)/(3x)#

#= 3 lim_(x to 0) sin(3x)/(3x)#

with sub #u = 3x#
#= 3 lim_(u to 0) sin(u)/(u)#

#=3#