How do you find the limit of #((sin^2)3x) / (x^2)# as x approaches 0?

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Answer 1 · 2016-08-31T16:52:47

#=9#

we will use well-known limit #lim_(z to 0) (sin z)/z = 1#. We need to do some manipulation to use it here

#lim_(x to 0) (sin^2 3x) / (x^2)#

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

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

and we factor out the constants
#=9 lim_(x to 0) (sin 3x) / (3x) (sin 3x) / (3x)#

with sub #z = 3x#, we have

#=9 lim_(z to 0) (sin z) / (z) (sin z) / (z)#

and the limit of the product is the product of the limits

#=9 lim_(z to 0) (sin z) / (z) lim_(z to 0) (sin z) / (z)#

#=9* 1* 1 = 9#