How do you find the limit of #(sin3x)/(2x)# as #x->0#?

2 Answers
Roy E.
Dec 23, 2016

#3/2#. By McLaurin Series for #sin 3x# and cancelling #x#.

Steve M
Dec 23, 2016

# lim_(x rarr 0) (sin3x)/(2x) = 3/2 #

Explanation:

An alternate proof:

# lim_(x rarr 0) (sin3x)/(2x) = lim_(x rarr 0) (sin3x)/(2x)*(3/2)/(3/2) #
# " " = lim_(x rarr 0) (3/2sin3x)/(3x) #
# " " = 3/2lim_(x rarr 0) (sin3x)/(3x) #
# " " = 3/2lim_(3x rarr 0) (sin3x)/(3x) " as " 3x rarr 0 " as " x rarr 0#
# " " = 3/2lim_(theta rarr 0) (sin theta)/(theta) " where " theta=3x#
# " " = 3/2 " as "lim_(theta rarr 0) (sin theta)/(theta) = 1#

Related questions
See all questions in Determining Limits Algebraically
Impact of this question
3255 views around the world
You can reuse this answer
Creative Commons License