2n-1+ax?+bx9. 设 f(x)=lim为连续函数，求a,b.x2n +1n-00解：当 |xl > 1 时，lim x2n-1= lim x2n = c0n-8n-8ba1+:1x2n-3 +x2n-2f(x) = lim1n-0xx+x2n-1当[xl<1时，limx2n-11 = limx2n = 0n-80n-80 +ax2 + bxf(x) ==ax2 + bx0+11+a+b当x = 1时，f(x) =2-1+a-b当x =-1时，f(x)=2099. 设 2 -1 2 2 + ( )=lim 1 n n n x ax bx f x → x + + 为连续函数，求𝑎, 𝑏 . 当 𝑥 < 1 时， 解: 𝑓 𝑥 = lim 𝑛→∞ 1 + 𝑎 𝑥 2𝑛−3 + 𝑏 𝑥 2𝑛−2 𝑥 + 1 𝑥 2𝑛−1 = 1 𝑥 当 𝑥 > 1 时， lim 𝑛→∞ 𝑥 2𝑛−1 = lim 𝑛→∞ 𝑥 2𝑛 = 0 lim 𝑛→∞ 𝑥 2𝑛−1 = lim 𝑛→∞ 𝑥 2𝑛 = ∞ 𝑓 𝑥 = 0 + 𝑎𝑥 2 + 𝑏𝑥 0 + 1 = 𝑎𝑥 2 + 𝑏𝑥 当𝑥 = 1 时，𝑓 𝑥 = 1 + 𝑎 + 𝑏 2 当𝑥 = −1 时，𝑓 𝑥 = −1 + 𝑎 − 𝑏 2 9