匿名
匿名 發問於 科學及數學數學 · 9 年 前

問以下e條limit of function

When x 不等於0, f(x) = (x^3)[sin (1/x)]

When x = 0, f(x) = 0

題目話Prove that f ' (0) exists.

但係

f ' (0) = lim h->0 [f(0+h)-f(0)]/h

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

As lim h->0 sin(1/h) does not exist

f ' (0) 應該都係does not exist架... 點解人會要人prove佢係exist....

佢marking用sandwich law 計到f ' (0) = 0

有冇人可以教下小弟?A level pastpaper來的....

2 個解答

評分
  • 蛙人
    Lv 6
    9 年 前
    最佳解答

    limf(x)g(x) = limf(x) limg(x) if the two limits on RHS are both exist.

    In this case, limsin(1/h) does not exist, so we can not deduce limh^2sin(1/h) does not exist.

    For details, please refer to the chapter about limit of function.

    2010-11-06 00:40:21 補充:

    簡單D講,例如 xsin(1/x) (x -> 0)

    x無論點變細, sin1/x既振幅只會係1 同- 1 之間

    但出面既x 愈來愈細,慢慢趨近0

    -> xsin(1/x)既振幅會由原來既1同-1,愈變愈細,最後振幅趨向0

    所以 limxsin(1/x) = 0

    資料來源: me
  • 匿名
    Lv 7
    9 年 前

    f ' (0) = lim h->0 [f(0+h)-f(0)]/h

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

    = lim h->0 h [sin(1/h)/(1/h)]

    = [lim h->0 h][lim h->0 sin(1/h)/(1/h)]

    = (lim h->0 h)(1)

    = 0*1

    = 0

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