PMaths 點show 佢 differentiable?

2009-12-26 1:47 am
f(x) = x^3 sin(1/x) if x<> 0
= 0 if x =0

要prove f'(0) exists 既話有兩個方法:
1. 第一個方法係做lim h tends to 0+ [f(x+h) - f(x)] / h =lim h tends to 0- [f(x+h) - f(x)] / h
但我想知用這個方法的話仲要唔要show f (0) 係continuous? 點解?

2. 第二個方法係show f'(0+) = f'(0-), 但我想知用這個方法的話仲要唔要show f (0) 係continuous? 點解? 如果f(0) continuous 而f'(0+) = f'(0-) = a, f' (0) 是否等於a?
更新1:

更正 f(x) = x^3 sin(1/x) if x 不等於0 = 0 if x =0 上面出現f'(0) 其實係devirative of f(0) 咁解

更新2:

最後更正: 要prove f'(0) exists 既話有兩個方法: 1. 第一個方法係做lim h tends to 0+ [f(x+h) - f(x)] / h =lim h tends to 0- [f(x+h) - f(x)] / h 但我想知用這個方法的話仲要唔要show f (0) 係continuous? 點解?

更新3:

2. 第二個方法係show f'(0+) = f'(0-), 但我想知用這個方法的話仲要唔要show f (0) 係continuous? 點解? 如果f(0) continuous 而f'(0+) = f'(0-) = a, f' (0) 是否等於a?

回答 (2)

2009-12-26 3:15 am
✔ 最佳答案
可導蘊含連續﹐所以根本不用想得這麼複雜。只要証明lim h -> 0 [f(0+h) - f(0)] / h 便行了。

lim h -> 0 [f(0+h) - f(0)] / h
=lim h -> 0 [(h)^3sin(1/h)-0]/h
=lim h -> 0 [(h)^2sin(1/h)]
=lim h->0 h[sin(1/h)/(1/h)]
=0

所以f'(0) exists
2010-02-07 6:40 pm
最佳解答中的証明出錯。
lim h -> 0 [f(0+h) - f(0)] / h
=lim h -> 0 [(h)^3sin(1/h)-0]/h
=lim h -> 0 [(h)^2sin(1/h)]
=lim h->0 h[sin(1/h)/(1/h)] 這一步有問題,h->0, 1/h -> infinity, 所以這不等於0
正確方法是直接使用 sine function bounded between -1 to 1 的事實做。
lim h -> 0 [(h)^2sin(1/h)] = 0 because |sin(1/h)| < 1.


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