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討論串[極限] x*lnx , x→0+
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lim x*lnx = ?. x→0+. 我看軟體是用L'Hospital做的. lim x*lnx. x→0+. lim lnx. = x→0+ ───. 1/x. L' lim 1/x. = x→0+ ───. -1/x^2. lim. = x→0+ (-x) = 0. 可是有個問題是. L'H
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Let lnx = -u. then x=e^{-u}. x→0+ becomes u→∞. --. ※ 發信站: 批踢踢實業坊(ptt.cc). ◆ From: 112.104.129.200. Forget it if you are not interested in mathematical
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if f(∞)=0,g(∞)=0. let x=1/u. lim_{x→∞} f(x)/g(x). = lim_{u→0+} f(1/u)/g(1/u). = lim_{u→0+} [(f(1/u) - 0)/(u-0)] / [(g(1/u) - 0)/(u-0)] --- (*). = [df(
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You can take f(1/u)=p(u) and g(1/u)=q(u). when u→0+, then p(u)→0, q(u)→0. Use ORDINARY HOSPITAL's RULE. lim_{u→0+} f(1/u)/g(1/u). = lim_{u→0+} p(u)/q(
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