【微分積分】3-5-3 不定形の極限｜要点まとめ
このページでは、大学数学・微分積分で学ぶ「不定形の極限」について、ロピタルの定理を用いた求め方を中心に解説します。定義・証明・例題を通して、不定形の扱い方と極限計算の本質を体系的に理解できます。
ロピタルの定理の内容と使い方
【ロピタルの定理】
関数\(f(x)\)と\(g(x)\)はともに点\(a\)のある近傍上で点\(a\)を除いて微分可能で、\(g'(x)\neq0\)とする。このとき
\(\displaystyle \lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0\)
または
\(\displaystyle \lim_{x\to a}f(x)=\lim_{x\to a}g(x)=\pm\infty\)
であり、さらに極限\(\displaystyle \lim_{x\to a}\frac{f'(x)}{g'(x)}\)が存在すれば、\(\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}\)も存在し
\(\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}\)
が成り立つ。
【例題】次の関数の極限値を求めなさい。
(1)\(\displaystyle \lim_{x\to0}\frac{x-\log(x+1)}{x^2}\)
前提条件より
\(\displaystyle \lim_{x\to0}\{x-\log(x+1)\}=0\)
\(\displaystyle \lim_{x\to0}x^2=0\)
\((x^2)'=2x\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to0}\frac{x-\log(x+1)}{x^2}\)
\(\displaystyle =\lim_{x\to0}\frac{\{x-\log(x+1)\}'}{(x^2)'}\)
\(\displaystyle =\lim_{x\to0}\frac{1-\frac{1}{x+1}}{2x}\)
\(\displaystyle =\lim_{x\to0}\frac{1}{2(x+1)}\)
\(\displaystyle =\frac{1}{2}\)
\(\displaystyle \lim_{x\to0}\{x-\log(x+1)\}=0\)
\(\displaystyle \lim_{x\to0}x^2=0\)
\((x^2)'=2x\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to0}\frac{x-\log(x+1)}{x^2}\)
\(\displaystyle =\lim_{x\to0}\frac{\{x-\log(x+1)\}'}{(x^2)'}\)
\(\displaystyle =\lim_{x\to0}\frac{1-\frac{1}{x+1}}{2x}\)
\(\displaystyle =\lim_{x\to0}\frac{1}{2(x+1)}\)
\(\displaystyle =\frac{1}{2}\)
(2)\(\displaystyle \lim_{x\to0}\frac{\tan x-x}{x-\sin x}\)
前提条件より
\(\displaystyle \lim_{x\to0}(\tan x-x)=0\)
\(\displaystyle \lim_{x\to0}(x-\sin x)=0\)
\((x-\sin x)'=1-\cos x\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to0}\frac{\tan x-x}{x-\sin x}\)
\(\displaystyle =\lim_{x\to0}\frac{(\tan x-x)'}{(x-\sin x)'}\)
\(\displaystyle =\lim_{x\to0}\frac{\frac{1}{\cos^2x}-1}{1-\cos x}\)
\(\displaystyle =\lim_{x\to0}\frac{1+\cos x}{\cos^2x}\)
\(=2\)
\(\displaystyle \lim_{x\to0}(\tan x-x)=0\)
\(\displaystyle \lim_{x\to0}(x-\sin x)=0\)
\((x-\sin x)'=1-\cos x\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to0}\frac{\tan x-x}{x-\sin x}\)
\(\displaystyle =\lim_{x\to0}\frac{(\tan x-x)'}{(x-\sin x)'}\)
\(\displaystyle =\lim_{x\to0}\frac{\frac{1}{\cos^2x}-1}{1-\cos x}\)
\(\displaystyle =\lim_{x\to0}\frac{1+\cos x}{\cos^2x}\)
\(=2\)
(3)\(\displaystyle \lim_{x\to\infty}x^2e^{-3x}\)
\(\displaystyle =\lim_{x\to\infty}\frac{x^2}{e^{3x}}\)
前提条件より
\(\displaystyle \lim_{x\to\infty}x^2=\infty\)
\(\displaystyle \lim_{x\to\infty}e^{3x}=\infty\)
\((e^{3x})'=3e^{3x}\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to\infty}\frac{x^2}{e^{3x}}\)
\(\displaystyle =\lim_{x\to\infty}\frac{(x^2)'}{(e^{3x})'}\)
\(\displaystyle =\lim_{x\to\infty}\frac{2x}{3e^{3x}}\)
前提条件より
\(\displaystyle \lim_{x\to\infty}2x=\infty\)
\(\displaystyle \lim_{x\to\infty}3e^{3x}=\infty\)
\((3e^{3x})'=9e^{3x}\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to\infty}\frac{2x}{3e^{3x}}\)
\(\displaystyle =\lim_{x\to\infty}\frac{(2x)'}{(3e^{3x})'}\)
\(\displaystyle =\lim_{x\to\infty}\frac{2}{9e^{3x}}\)
\(=0\)
前提条件より
\(\displaystyle \lim_{x\to\infty}x^2=\infty\)
\(\displaystyle \lim_{x\to\infty}e^{3x}=\infty\)
\((e^{3x})'=3e^{3x}\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to\infty}\frac{x^2}{e^{3x}}\)
\(\displaystyle =\lim_{x\to\infty}\frac{(x^2)'}{(e^{3x})'}\)
\(\displaystyle =\lim_{x\to\infty}\frac{2x}{3e^{3x}}\)
前提条件より
\(\displaystyle \lim_{x\to\infty}2x=\infty\)
\(\displaystyle \lim_{x\to\infty}3e^{3x}=\infty\)
\((3e^{3x})'=9e^{3x}\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to\infty}\frac{2x}{3e^{3x}}\)
\(\displaystyle =\lim_{x\to\infty}\frac{(2x)'}{(3e^{3x})'}\)
\(\displaystyle =\lim_{x\to\infty}\frac{2}{9e^{3x}}\)
\(=0\)
(4)\(\displaystyle \lim_{x\to0}\frac{x-\sin x}{x^3}\)
前提条件より
\(\displaystyle \lim_{x\to0}(x-\sin x)=0\)
\(\displaystyle \lim_{x\to0}x^3=0\)
\((x^3)'=3x^2\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to0}\frac{x-\sin x}{x^3}\)
\(\displaystyle =\lim_{x\to0}\frac{(x-\sin x)'}{(x^3)'}\)
\(\displaystyle =\lim_{x\to0}\frac{1-\cos x}{3x^2}\)
前提条件より
\(\displaystyle \lim_{x\to0}(1-\cos x)=0\)
\(\displaystyle \lim_{x\to0}3x^2=0\)
\((3x^2)'=6x\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to0}\frac{1-\cos x}{3x^2}\)
\(\displaystyle =\lim_{x\to0}\frac{(1-\cos x)'}{(3x^2)'}\)
\(\displaystyle =\lim_{x\to0}\frac{\sin x}{6x}\)
\(\displaystyle =\frac{1}{6}\)
\(\displaystyle \lim_{x\to0}(x-\sin x)=0\)
\(\displaystyle \lim_{x\to0}x^3=0\)
\((x^3)'=3x^2\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to0}\frac{x-\sin x}{x^3}\)
\(\displaystyle =\lim_{x\to0}\frac{(x-\sin x)'}{(x^3)'}\)
\(\displaystyle =\lim_{x\to0}\frac{1-\cos x}{3x^2}\)
前提条件より
\(\displaystyle \lim_{x\to0}(1-\cos x)=0\)
\(\displaystyle \lim_{x\to0}3x^2=0\)
\((3x^2)'=6x\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to0}\frac{1-\cos x}{3x^2}\)
\(\displaystyle =\lim_{x\to0}\frac{(1-\cos x)'}{(3x^2)'}\)
\(\displaystyle =\lim_{x\to0}\frac{\sin x}{6x}\)
\(\displaystyle =\frac{1}{6}\)
(5)\(\displaystyle \lim_{x\to\frac{\pi}{2}}\frac{e^{\sin x}-e}{\log(\sin x)}\)
前提条件より
\(\displaystyle \lim_{x\to\frac{\pi}{2}}(e^{\sin x}-e)=0\)
\(\displaystyle \lim_{x\to\frac{\pi}{2}}\log(\sin x)=0\)
\(\displaystyle (\log(\sin x))'=\frac{\cos x}{\sin x}\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to\frac{\pi}{2}}\frac{e^{\sin x}-e}{\log(\sin x)}\)
\(\displaystyle =\lim_{x\to\frac{\pi}{2}}\frac{(e^{\sin x}-e)'}{(\log(\sin x))'}\)
\(\displaystyle =\lim_{x\to\frac{\pi}{2}}\frac{e^{\sin x}\cos x}{\frac{\cos x}{\sin x}}\)
\(=e\)
\(\displaystyle \lim_{x\to\frac{\pi}{2}}(e^{\sin x}-e)=0\)
\(\displaystyle \lim_{x\to\frac{\pi}{2}}\log(\sin x)=0\)
\(\displaystyle (\log(\sin x))'=\frac{\cos x}{\sin x}\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to\frac{\pi}{2}}\frac{e^{\sin x}-e}{\log(\sin x)}\)
\(\displaystyle =\lim_{x\to\frac{\pi}{2}}\frac{(e^{\sin x}-e)'}{(\log(\sin x))'}\)
\(\displaystyle =\lim_{x\to\frac{\pi}{2}}\frac{e^{\sin x}\cos x}{\frac{\cos x}{\sin x}}\)
\(=e\)
(6)\(\displaystyle \lim_{x\to\infty}\frac{\log x}{\sqrt{x}}\)
前提条件より
\(\displaystyle \lim_{x\to\infty}\log x=\infty\)
\(\displaystyle \lim_{x\to\infty}\sqrt{x}=\infty\)
\(\displaystyle (\sqrt{x})'=\frac{1}{2\sqrt{x}}\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to\infty}\frac{\log x}{\sqrt{x}}\)
\(\displaystyle =\lim_{x\to\infty}\frac{(\log x)'}{(\sqrt{x})'}\)
\(\displaystyle =\lim_{x\to\infty}\frac{\frac{1}{x}}{\frac{1}{2\sqrt{x}}}\)
\(=0\)
\(\displaystyle \lim_{x\to\infty}\log x=\infty\)
\(\displaystyle \lim_{x\to\infty}\sqrt{x}=\infty\)
\(\displaystyle (\sqrt{x})'=\frac{1}{2\sqrt{x}}\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to\infty}\frac{\log x}{\sqrt{x}}\)
\(\displaystyle =\lim_{x\to\infty}\frac{(\log x)'}{(\sqrt{x})'}\)
\(\displaystyle =\lim_{x\to\infty}\frac{\frac{1}{x}}{\frac{1}{2\sqrt{x}}}\)
\(=0\)
(7)\(\displaystyle \lim_{x\to+0}x\log x\)
\(\displaystyle =\lim_{x\to+0}\frac{\log x}{x^{-1}}\)
前提条件より
\(\displaystyle \lim_{x\to+0}\log x=-\infty\)
\(\displaystyle \lim_{x\to+0}x^{-1}=+\infty\)
\((x^{-1})'=-x^{-2}\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to+0}\frac{\log x}{x^{-1}}\)
\(\displaystyle =\lim_{x\to+0}\frac{(\log x)'}{(x^{-1})'}\)
\(\displaystyle =\lim_{x\to+0}\frac{\frac{1}{x}}{-x^{-2}}\)
\(=0\)
前提条件より
\(\displaystyle \lim_{x\to+0}\log x=-\infty\)
\(\displaystyle \lim_{x\to+0}x^{-1}=+\infty\)
\((x^{-1})'=-x^{-2}\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to+0}\frac{\log x}{x^{-1}}\)
\(\displaystyle =\lim_{x\to+0}\frac{(\log x)'}{(x^{-1})'}\)
\(\displaystyle =\lim_{x\to+0}\frac{\frac{1}{x}}{-x^{-2}}\)
\(=0\)
(8)\(\displaystyle \lim_{x\to\infty}x\log\frac{x+1}{x-1}\)
\(\displaystyle =\lim_{x\to\infty}\frac{\log\frac{x+1}{x-1}}{x^{-1}}\)
前提条件より
\(\displaystyle \lim_{x\to\infty}\log\frac{x+1}{x-1}=0\)
\(\displaystyle \lim_{x\to\infty}x^{-1}=0\)
\((x^{-1})'=-x^{-2}\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to\infty}\frac{\log\frac{x+1}{x-1}}{x^{-1}}\)
\(\displaystyle =\lim_{x\to\infty}\frac{(\log\frac{x+1}{x-1})'}{(x^{-1})'}\)
\(\displaystyle =\lim_{x\to\infty}\frac{\frac{1}{x+1}-\frac{1}{x-1}}{-x^{-2}}\)
\(=2\)
前提条件より
\(\displaystyle \lim_{x\to\infty}\log\frac{x+1}{x-1}=0\)
\(\displaystyle \lim_{x\to\infty}x^{-1}=0\)
\((x^{-1})'=-x^{-2}\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to\infty}\frac{\log\frac{x+1}{x-1}}{x^{-1}}\)
\(\displaystyle =\lim_{x\to\infty}\frac{(\log\frac{x+1}{x-1})'}{(x^{-1})'}\)
\(\displaystyle =\lim_{x\to\infty}\frac{\frac{1}{x+1}-\frac{1}{x-1}}{-x^{-2}}\)
\(=2\)
(9)\(\displaystyle \lim_{x\to1}\left(\frac{1}{x-1}-\frac{1}{\log x}\right)\)
\(\displaystyle =\lim_{x\to1}\frac{\log x-x+1}{(x-1)\log x}\)
前提条件より
\(\displaystyle \lim_{x\to1}(\log x-x+1)=0\)
\(\displaystyle \lim_{x\to1}(x-1)\log x=0\)
\(\displaystyle ((x-1)\log x)'=\log x+\frac{x-1}{x}\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to1}\frac{\log x-x+1}{(x-1)\log x}\)
\(\displaystyle =\lim_{x\to1}\frac{(\log x-x+1)'}{\{(x-1)\log x\}'}\)
\(\displaystyle =\lim_{x\to1}\frac{\frac{1}{x}-1}{\log x+\frac{x-1}{x}}\)
\(\displaystyle =\lim_{x\to1}\frac{1-x}{x\log x+x-1}\)
前提条件より
\(\displaystyle \lim_{x\to1}(1-x)=0\)
\(\displaystyle \lim_{x\to1}(x\log x+x-1)=0\)
\((x\log x+x-1)'=\log x+2\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to1}\frac{1-x}{x\log x+x-1}\)
\(\displaystyle =\lim_{x\to1}\frac{(1-x)'}{(x\log x+x-1)'}\)
\(\displaystyle =\lim_{x\to1}\frac{-1}{\log x+2}\)
\(\displaystyle =-\frac{1}{2}\)
前提条件より
\(\displaystyle \lim_{x\to1}(\log x-x+1)=0\)
\(\displaystyle \lim_{x\to1}(x-1)\log x=0\)
\(\displaystyle ((x-1)\log x)'=\log x+\frac{x-1}{x}\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to1}\frac{\log x-x+1}{(x-1)\log x}\)
\(\displaystyle =\lim_{x\to1}\frac{(\log x-x+1)'}{\{(x-1)\log x\}'}\)
\(\displaystyle =\lim_{x\to1}\frac{\frac{1}{x}-1}{\log x+\frac{x-1}{x}}\)
\(\displaystyle =\lim_{x\to1}\frac{1-x}{x\log x+x-1}\)
前提条件より
\(\displaystyle \lim_{x\to1}(1-x)=0\)
\(\displaystyle \lim_{x\to1}(x\log x+x-1)=0\)
\((x\log x+x-1)'=\log x+2\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to1}\frac{1-x}{x\log x+x-1}\)
\(\displaystyle =\lim_{x\to1}\frac{(1-x)'}{(x\log x+x-1)'}\)
\(\displaystyle =\lim_{x\to1}\frac{-1}{\log x+2}\)
\(\displaystyle =-\frac{1}{2}\)
(10)\(\displaystyle \lim_{x\to0}\left(\frac{1}{x^2}-\frac{1}{x\sin x}\right)\)
\(\displaystyle =\lim_{x\to0}\frac{\sin x-x}{x^2\sin x}\)
前提条件より
\(\displaystyle \lim_{x\to0}(\sin x-x)=0\)
\(\displaystyle \lim_{x\to0}x^2\sin x=0\)
\((x^2\sin x)'=2x\sin x+x^2\cos x\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to0}\frac{\sin x-x}{x^2\sin x}\)
\(\displaystyle =\lim_{x\to0}\frac{(\sin x-x)'}{(x^2\sin x)'}\)
\(\displaystyle =\lim_{x\to0}\frac{\cos x-1}{2x\sin x+x^2\cos x}\)
前提条件より
\(\displaystyle \lim_{x\to0}(\cos x-1)=0\)
\(\displaystyle \lim_{x\to0}(2x\sin x+x^2\cos x)=0\)
\((2x\sin x+x^2\cos x)'\)
\(\ \ \ \ \ =(2-x^2)\sin x+4x\cos x\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to0}\frac{\cos x-1}{2x\sin x+x^2\cos x}\)
\(\displaystyle =\lim_{x\to0}\frac{(\cos x-1)'}{(2x\sin x+x^2\cos x)'}\)
\(\displaystyle =\lim_{x\to0}\frac{-\sin x}{(2-x^2)\sin x+4x\cos x}\)
\(\displaystyle =\lim_{x\to0}\frac{-1}{(2-x^2)+4\cos x\frac{x}{\sin x}}\)
\(\displaystyle =-\frac{1}{6}\)
前提条件より
\(\displaystyle \lim_{x\to0}(\sin x-x)=0\)
\(\displaystyle \lim_{x\to0}x^2\sin x=0\)
\((x^2\sin x)'=2x\sin x+x^2\cos x\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to0}\frac{\sin x-x}{x^2\sin x}\)
\(\displaystyle =\lim_{x\to0}\frac{(\sin x-x)'}{(x^2\sin x)'}\)
\(\displaystyle =\lim_{x\to0}\frac{\cos x-1}{2x\sin x+x^2\cos x}\)
前提条件より
\(\displaystyle \lim_{x\to0}(\cos x-1)=0\)
\(\displaystyle \lim_{x\to0}(2x\sin x+x^2\cos x)=0\)
\((2x\sin x+x^2\cos x)'\)
\(\ \ \ \ \ =(2-x^2)\sin x+4x\cos x\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to0}\frac{\cos x-1}{2x\sin x+x^2\cos x}\)
\(\displaystyle =\lim_{x\to0}\frac{(\cos x-1)'}{(2x\sin x+x^2\cos x)'}\)
\(\displaystyle =\lim_{x\to0}\frac{-\sin x}{(2-x^2)\sin x+4x\cos x}\)
\(\displaystyle =\lim_{x\to0}\frac{-1}{(2-x^2)+4\cos x\frac{x}{\sin x}}\)
\(\displaystyle =-\frac{1}{6}\)
(11)\(\displaystyle \lim_{x\to+0}x^{\sqrt{x}}\)
対数をとると
\(\displaystyle \lim_{x\to+0}\frac{\log x}{\frac{1}{\sqrt{x}}}\)
前提条件より
\(\displaystyle \lim_{x\to+0}\log x=-\infty\)
\(\displaystyle \lim_{x\to+0}\frac{1}{\sqrt{x}}=+\infty\)
\(\displaystyle \left(\frac{1}{\sqrt{x}}\right)'=-\frac{1}{2x\sqrt{x}}\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to+0}\frac{\log x}{\frac{1}{\sqrt{x}}}\)
\(\displaystyle =\lim_{x\to+0}\frac{(\log x)'}{(\frac{1}{\sqrt{x}})'}\)
\(\displaystyle =\lim_{x\to+0}\frac{\frac{1}{x}}{-\frac{1}{2x\sqrt{x}}}\)
\(=0\)
\(\displaystyle \lim_{x\to+0}\frac{\log x}{\frac{1}{\sqrt{x}}}\)
前提条件より
\(\displaystyle \lim_{x\to+0}\log x=-\infty\)
\(\displaystyle \lim_{x\to+0}\frac{1}{\sqrt{x}}=+\infty\)
\(\displaystyle \left(\frac{1}{\sqrt{x}}\right)'=-\frac{1}{2x\sqrt{x}}\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to+0}\frac{\log x}{\frac{1}{\sqrt{x}}}\)
\(\displaystyle =\lim_{x\to+0}\frac{(\log x)'}{(\frac{1}{\sqrt{x}})'}\)
\(\displaystyle =\lim_{x\to+0}\frac{\frac{1}{x}}{-\frac{1}{2x\sqrt{x}}}\)
\(=0\)
(12)\(\displaystyle \lim_{x\to\infty}\left(\frac{2}{\pi}\tan^{-1}x\right)^x\)
対数をとると
\(\displaystyle \lim_{x\to\infty}\frac{\log(\frac{2}{\pi}\tan^{-1}x)}{\frac{1}{x}}\)
前提条件より
\(\displaystyle \lim_{x\to\infty}\log\left(\frac{2}{\pi}\tan^{-1}x\right)=0\)
\(\displaystyle \lim_{x\to\infty}\frac{1}{x}=0\)
\(\displaystyle \left(\frac{1}{x}\right)'=-x^{-2}\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to\infty}\frac{\log(\frac{2}{\pi}\tan^{-1}x)}{\frac{1}{x}}\)
\(\displaystyle =\lim_{x\to\infty}\frac{\{\log(\frac{2}{\pi}\tan^{-1}x)\}'}{(\frac{1}{x})'}\)
\(\displaystyle =\lim_{x\to\infty}\frac{\frac{1}{\tan^{-1}x}・\frac{1}{1+x^2}}{-\frac{1}{x^2}}\)
\(\displaystyle =-\frac{2}{\pi}\)
\(\displaystyle \lim_{x\to\infty}\frac{\log(\frac{2}{\pi}\tan^{-1}x)}{\frac{1}{x}}\)
前提条件より
\(\displaystyle \lim_{x\to\infty}\log\left(\frac{2}{\pi}\tan^{-1}x\right)=0\)
\(\displaystyle \lim_{x\to\infty}\frac{1}{x}=0\)
\(\displaystyle \left(\frac{1}{x}\right)'=-x^{-2}\neq0\)
ロピタルの定理より
\(\displaystyle \lim_{x\to\infty}\frac{\log(\frac{2}{\pi}\tan^{-1}x)}{\frac{1}{x}}\)
\(\displaystyle =\lim_{x\to\infty}\frac{\{\log(\frac{2}{\pi}\tan^{-1}x)\}'}{(\frac{1}{x})'}\)
\(\displaystyle =\lim_{x\to\infty}\frac{\frac{1}{\tan^{-1}x}・\frac{1}{1+x^2}}{-\frac{1}{x^2}}\)
\(\displaystyle =-\frac{2}{\pi}\)
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