I will run through two methods: the 'unofficial' method, and the (more) 'official' method.
Intuitive approach around the limit:
Using the compound angle and double angle formulas for #sin# and #cos#:
#sin(3y) = sin (2y + y)#
#=sin(2y)cosy + cos(2y)siny)#
#=2sinycos^2y+(2cos^2y-1)siny#
#=4sinycos^2y - siny#
(Rearranging to get it all in terms of #sin(y)#)
#=4siny(1-sin^2y) - siny#
#=3siny - 4sin^3y#
Now we want to find the square of that (#sin(3y)#):
#sin^2(3y) = (3siny - 4sin^3y)^2#
#=(siny(3-4sin^2y))^2#
#=sin^2y(3- (4 - 4cos^2y))^2#
#= sin^2y(4cos^2y - 1)^2#
Now to find:
#lim_(y rarr 0) (sin^2y(4cos^2y - 1)^2 - 3y^2)/(y^2 + y^3)#
This is where we start to get a bit 'intuitive'...
As #y rarr 0#, #siny rarr 0#
#rArr siny rarr y#
Hence let #y = siny#.
Meanwhile #cosy rarr 1#
Hence let #cosy = 1#.
This simplifies the limit down to:
#lim_(y rarr 0) (y^2(3)^2 - 3y^2)/(y^2 + y^3)#
#=lim_(y rarr 0)(6y^2)/(y^2 + y^3)#
#=6 lim_(y rarr 0) y^2/(y^2(1+y))#
#=6 lim_(y rarr 0) 1/(1+y)#
Finally letting #y = 0# we get:
#6 lim_(y rarr 0) 1#
#= 6#
Official approach to the limit:
This uses L'Hopital's Rule, i.e. if #lim_(y rarr 0) f(x)/g(x)# yields a mathematically impossible result, then the limit will be the same if you repeatedly differentiate #f# and #g# to the point where the result is no longer undefined.
#lim_(y rarr 0) (sin^2(3y) - 3y^2)/(y^2 + y^3)#
#lim_(y rarr 0) (d/dx(sin^2(3y) - 3y^2))/(d/dx(y^2 + y^3))#
By the chain rule, this is equal to:
#lim_(y rarr 0) (6sin(3y)cos(3y) - 6y)/(2y + 3y^2)#
Still gives out the zero denominator, so we do it all again!
By the product and chain rules:
#d/dx(6sin(3y)cos(3y)#
#=6(3cos(3y)cos(3y) + (-sin(3y)sin(3y))#
#=6(3cos^3y - 3sin^3y)#
Hence:
#lim_(y rarr 0) (6sin(3y)cos(3y) - 6y)/(2y + 3y^2)#
#=lim_(y rarr 0) (18cos^2(3y) - 18sin^2(3y) - 6)/(2 +6y)#
Rearranging the trigonometric parts of the numerator using the Pythagorean Identity, we get:
#lim_(y rarr 0) (36cos^2(3y) - 18 - 6)/(2 +6y)#
Finally letting #y=0#, we get:
#(36-18-6)/2#
#=6#
Goodness, that was long - I hope it helps!
