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※ 引述《henry9621205 (清心小子)》之銘言: : {ln[x^(1/5)]} (sinx)^2 : lim----------------------- = ? : x→0 x^(1/2) lim (sinx / x)^2 * (1/5) * x^(3/2) * ln(x) = 0 (其中 lim x^(3/2) * lnx = 0 當羅必達練習題) : lim[(π/2-x)ln|tanx|] = ? : x→π/2 左極限 α = π/2 - x > 0 lim -αln(tanα) α-> 0 = lim -α[ln(tanα /α) + lnα] = 0 右極限給你當練習 左極限 = 右極限 = 0 : 先說聲謝了! -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.109.103.226
legstrong:第二題tanx怎麼變成tanα的阿? 112.105.123.85 05/18 18:09
Honor1984:tan(π/2 - α) = 1/tanα140.109.103.226 05/18 18:16
henry9621205:謝謝 114.41.50.164 05/20 00:16