Question

高等数学求极限，2，12怎么做？

Answer 1

2.12 f(x) 即为分段函数：
f(x) = x^2 x > 1；
f(x) = (1/2)(1+a+b) x = 1 ;
f(x) = ax+b x < 1 .
函数要连续，则
(1/2)(1+a+b) = 1， (1/2)(1+a+b) = a+b
得 a+b = 1

追问

能问下你分段函数怎么求的吗？我就是这个不懂

追答

因有 e^[n(x-1)] 项， 分 x > 1, x = 1, x 1 时，lime^[n(x-1)] = + ∞,
分子分母同乘以 e^[-n(x-1)]， 则 lime^[-n(x-1)] = 0,
f(x) = lim{x^2 + (ax+b)e^[-n(x-1)]}/{e^[-n(x-1)]+1} = x^2;
当 x = 1 时, f(x) = lim(1+a+b)/2 = (1/2)(1+a+b);
当 x e^[n(x-1)] = 0,
f(x) = lim{x^2e^[n(x-1)] + ax+b}/{1+e^[n(x-1)]} = ax+b.