例 已知 limax+b)=3,求常数a、b.x+1x-8+(b-a)x+1+b(1-a)解原极限=lim3x+1x-→001-a=0a=b-a=3b= 4
6 例 ) 3, 1 1 lim ( 2 − + = + + → a x b x x x 已 知 求 常 数a、b. 解 原极限= 1 (1 ) ( ) 1 lim 2 + − + − + + → x a x b a x b x = 3 1− a = 0 b−a = 3 a = 1 b = 4
2n-1+ ax + bX例 设f(x) = lim为连续函数x2n +1n8求a、b.ax+bIxk1解1[x[>1x：f(x)连续f(x) =Y(1+a+b)x=121x=-1(-1-a+b):. x =-1时,f(-1-0)= f(-1+0) = f(-1) 即-1= -a + b =-1-a+b)7
7 例 , 1 ( ) lim 2 2 1 设 为连续函数 + + + = − → n n n x x a x b f x 求a、b. 解 f (x) = | x | 1 | x | 1 x = 1 x = −1 ax + b x1 (1 ) 21 + a + b ( 1 ) 21 − − a + b f ( x )连续, f ( − 1 − 0 ) = f ( − 1 + 0 ) = f ( − 1 ) 即 − 1 = − a + b ( 1 ) 21 = − − a + b x = − 1 时
1xk1ax+b-1=-a + b :-1-a+ b[x>1：f(x)连续f(x)=x=1(1+a+b)(-1-a+b)x=-12即:. x =1时, f(1-0) = f(1+ 0) = f(1)a + b=1 = =(1+ a+ b)得 b=0,a=18
8 f ( x ) = | x | 1 | x | 1 x = 1 x = − 1 ax + b x1 (1 ) 21 + a + b ( 1 ) 21 − − a + b f ( x )连续 , f ( 1 − 0 ) = f ( 1 + 0 ) = f ( 1 ) 即 a + b = 1 ( 1 ) 21 = + a + b x = 1 时 , − 1 = − a + b ( 1 ) 21 = − − a + b 得 b = 0 , a = 1