x? +ax+b6. 设 lim3,确定a、b的值x-1 sin(x2 -1)解：由于 limsin(x2-1)=0,故 lim(x? +ax+b)=1+a+b=0,即 b=-l-α，于是x? +ax + bx2-1+a(x-1)lim-= lim→1 sin(x? -1)x2 - 1x-ia= lim(1+ -x+1)x-1=1+α= 3解得 a=4,b=-52.66. 设 2 2 1 lim 3, sin( 1) x x ax b → x + + = − 确定 a、b 的值. 解: 6 = lim 𝑥→1 𝑥 ) 2 − 1 + 𝑎(𝑥 − 1 𝑥 2 − 1 = lim 𝑥→1 (1+ 𝑎 𝑥 + 1 ) 解得 a = 4,b = -5. 2 2 1 lim sin( 1) x x ax b → x + + − 即 b a = − −1 , 于是 故 2 1 lim( ) 1 =0， x x ax b a b → + + = + + 由于 2 1 limsin( 1) 0 x x → − = ， =1+ 1 2 𝑎 = 3