limx�¨0 sin ^3 (3x)/ x sin(x ^2)

Answer 1

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