2019年10月30日水曜日

学習環境

解析入門(上) (松坂和夫 数学入門シリーズ 4) (松坂 和夫(著)、岩波書店)の第6章(関数の近似、テイラーの定理)、6.2(極限の計算)、問題2の解答を求めてみる。



    1. lim x 0 x - log 1 + x = 0 lim x 0 x 2 = 0 d dx x - log 1 + x = 1 - 1 1 + x = x 1 + x d dx x 2 = 2 x lim x 0 x 1 + x · 1 2 x = 1 2

      よって、 ロピタルの 定理より

      lim x 0 x - log 1 + x x 2 = 1 2

    2. lim x 0 1 sin 2 x - 1 x 2 = lim x 0 x 2 - sin 2 x x 2 sin 2 x = lim x 0 x 2 sin 2 x · x 2 - sin 2 x x 4 lim x 0 x 2 - sin 2 x = 0 lim x 0 x 4 = 0 d dx x 2 - sin 2 x = 2 x - sin 2 x d dx x 4 = 4 x 3 lim x 0 2 x - sin 2 x = 0 lim x 0 4 x 3 = 0 d dx 2 x - sin 2 x = 2 - 2 cos 2 x d dx 4 x 3 = 12 x 2 lim x 0 2 - 2 cos 2 x = 0 lim x 0 12 x 2 = 0 d dx 2 - 2 cos 2 x = 4 sin 2 x d dx 12 x 2 = 24 x lim x 0 4 sin 2 x 24 x = 1 3 lim x 0 sin 2 x 2 x = 1 3 lim x 0 1 sin 2 x - 1 x 2 = 1 3

    3. lim x 0 x - sin x = 0 lim x 0 tan x - x = 0 d dx x - sin x = 1 - cos x d dx tan x - x = 1 cos 2 x - 1 = 1 - cos 2 x cos 2 x = 1 + cos x 1 - cos x cos 2 x lim x 0 1 - cos x · cos 2 x 1 + cos x 1 - cos x = 1 2 lim x 0 x - sin x tan x - x = 1 2

    4. lim x + x log x x 1 x - 1 = lim x + x 1 x - 1 log x 1 x lim x + log x 1 x = lim x + log x x lim x + 0 x = + d dx log x = 1 x d dx x = 1 lim x + 1 x = 0 lim x + log x 1 x = 0 lim x + x 1 x = 1 lim x + x log x x 1 x - 1 = lim y 1 y - 1 log y lim y 1 y - 1 = 0 lim y 1 log y = 0 d dy y - 1 = 1 d dy log y = 1 y lim y 1 1 1 y = 1 lim x + x log x x 1 x - 1 = 1

    5. lim x 0 + log 1 sin x x = lim x 0 + x log 1 sin x = lim x 0 + x sin x sin x log 1 sin x = lim x 0 + x sin x · log 1 sin x 1 sin x lim y + y = + d dy log y = 1 y d dy y = 1 lim y + 1 y = 0 lim y + log y y = 0 lim x 0 + log 1 sin x 1 sin x = 0 lim x 0 + log 1 sin x x = 0 lim x 0 + 1 sin x x = 1

    6. log 1 + x = x - 1 2 x 2 + ε

      とおくと、

      lim x 0 ε x 2 = 0

      が成り立つ。

      よって、

      log 1 + x 1 x = 1 x log 1 + x = 1 x x - 1 2 x 2 + ε = 1 - 1 2 x + ε x lim x 0 log 1 + x 1 x - 1 x = lim x 0 - 1 2 + ε x 2 = - 1 2 lim x 0 e - 1 + x 1 x x = lim x 0 log 1 + x 1 x - 1 x · e - 1 + x 1 x log 1 + x 1 x - 1 = lim x 0 log 1 + x 1 x - 1 x · e - e log 1 + x 1 x log 1 + x 1 x - 1 = lim x 0 log 1 + x 1 x - 1 x · e · 1 - exp log 1 + x 1 x - 1 log 1 + x 1 x - 1 lim x 0 log 1 + x 1 x = lim x 0 log 1 + x x lim x 0 log 1 + x = 0 lim x 0 x = 0 d dx log 1 + x = 1 1 + x d dx x = 1 lim x 0 1 1 + x = 1 lim x 0 log 1 + x 1 x = 1 lim y 1 1 - e y - 1 = 0 lim y 1 y - 1 = 0 d dy 1 - e y - 1 = - e y - 1 d dy y - 1 = - 1 lim y 1 e y - 1 = - 1 lim y 1 1 - e y - 1 y - 1 = - 1 lim x 0 1 - exp log 1 + x 1 x - 1 log 1 + x 1 x - 1 = - 1 lim x 0 e - 1 + x 1 x x = - 1 2 e - 1 = e 2

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