數學 M2 極限 急！！?

Let u = 3x
If x → 0 , then u → 0

as u → 0 ,
( sin 6x + 2x ) / ( sin 3x - 7x )
= [ sin 2u + (2/3)u ] / [ sin u - (7/3)u ]
= [ 2*sin u*cos u + (2/3)u ] / [ sin u - (7/3)u ]
= [ 2*( sin u / u )*cos u + 2/3 ] / [ ( sin u / u ) - 7/3 ] , note that u ≠ 0 , for u → 0
→ ( 2*1*1 + 2/3 ) / ( 1 - 7/3 ) , because sin u / u → 1
= (8/3) / (-4/3)
= - 2 ..... Ans