limx�¨0 sin ^3 (3x)/ x sin(x ^2)
If your expression mean:
lim sin ^ 3 ( 3 x ) / ( x sin ( x ^ 2) ) , as x -> 0
then :
f ( x ) = sin ^ 3 ( 3 x )
g ( x ) = x sin ( x ^ 2)
lim [ f ( x ) / g ( x ) ] as x->0 =
lim f ( x ) , as x -> 0 / lim g ( x ) , as x -> 0 =
lim sin ^ 3 ( 3 x ) , as x -> 0 / lim x sin ( x ^ 2 ) , as x -> 0
= 0 / 0
Now you must use L'Hôpital's rule :
lim f ( x ) , as x -> 0 / lim g ( x ) , as x -> 0 =
lim f´ ( x ) , as x -> 0 / lim g´ ( x ) , as x -> 0
f´ ( x ) = [ sin ^ 3 ( 3 x ) ] ´ = 9 sin ^ 2 ( 3 x ) * cos ( 3 x )
g´ ( x ) = [ x sin ( x ^ 2 ) ] ´ = sin ( x ^ 2 ) + 2 x ^ 2 * cos ( x ^ 2 )
lim f´ ( x ) , as x -> 0 / lim g´ ( x ) , as x -> 0 =
lim [ 9 sin ^ 2 ( 3 x ) * cos ( 3 x ) , as x ->0 ] / lim [ sin ( x ^ 2 ) * 2 x ^ 2 * cos ( x ^ 2 ) , as x ->0 ] =
0 / 0
Again L'Hôpital's rule :
lim f ( x ) , as x -> 0 / lim g ( x ) , as x -> 0 =
lim f " ( x ) , as x -> 0 / lim g " ( x ) , as x -> 0
f " ( x ) = 54 * sin ( 3 x ) * cos ^ 2 ( 3 x ) - 27 * sin ^ 3 ( 3 x )
g " ( x ) = 6 x * cos ( x ^ 2 ) - 4 x ^ 3 * sin ( x ^ 2 )
lim f " ( x ) , as x ->0 =
lim 54 * sin ( 3 x ) * cos ^ 2 ( 3 x ) - 27 * sin ^ 3 ( 3 x ) as x -> 0 = 0
lim g " ( x ) , as x -> 0 =
lim 6 x * cos ( x ^ 2 ) - 4 x ^ 3 * sin ( x ^ 2 ) as x -> 0 = 0
lim f " ( x ) , as x -> 0 / lim g " ( x ) , as x -> 0 = 0 / 0
Again L'Hôpital's rule :
lim f ( x ) , as x -> 0 / lim g ( x ) , as x -> 0 =
lim f ´´´ ( x ) , as x -> 0 / lim g ´´´ ( x ) , as x -> 0
f ´´´ ( x ) = 162 * cos ^ 3 ( 3 x ) - 567 * sin ^ 2 ( 3 x ) * cos ( 3 x )
g ´´´ ( x ) = ( 6 - 8 x ^ 4 ) * cos ( x ^ 2 ) - 24 x ^ 2 * sin ( x ^ 2 )
lim f ´´´ ( x ) , as x -> 0 =
lim 162 * cos ^ 3 ( 3 x ) - 567 * sin ^ 2 ( 3 x ) * cos ( 3 x ) as x -> 0 =
162
lim g ´´´ ( x ) , as x -> 0 =
lim ( 6 - 8 x ^ 4 ) * cos ( x ^ 2 ) - 24 x ^ 2 * sin ( x ^ 2 ) as x - > 0 =
6
lim sin ^ 3 ( 3 x ) / ( x sin ( x ^ 2) ) , as x -> 0 =
lim f ´´´ ( x ) , as x -> 0 / lim g ´´´ ( x ) , as x -> 0 = 162 / 6 = 27