分段函数求连续方便的话可以请您手写上传解题过程的截图吗?我今年考研,希望您能帮我解答
1个回答

1）……,a = f'(0)；

2）因

lim(x→0)[g(x) - g(0)]/x

= lim(x→0)[f(x)/x - f'(0)]/x

= lim(x→0)[f(x) - xf'(0)]/x^2 (0/0,用L'Hospital法则)

= lim(x→0)[f'(x) - f'(0)]/2x

= f"(0)/2,

得知

g'(x) = [xf'(x) - f(x)]/x^2,x ≠ 0,

= (1/2)f"(0),x = 0,

明显的,在 x ≠ 0,g'(x) 是连续的；而在 x = 0 处,

lim(x→0)g'(x)

= lim(x→0)[xf'(x) - f(x)]/x^2

= lim(x→0)[f'(x) - f'(0)]/x - lim(x→0)[f(x) - xf'(0)]/x^2

= f"(0) - f"(0)/2

= f"(0)/2,

得证.

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