%2007/04/22 06:10
%Copyright 2007 WATANABE,Masayuki
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\def\dsum{\displaystyle\sum}
\def\dlim{\displaystyle\lim}
\def\dint{\displaystyle\int}
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\begin{document}
{\large 京都大学　1972年　入学試験　理系数学　問題2}\\
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{\large 問題}\\
\\
　次の式で定められる関数 $F(x)$ に対して，$\dlim_{x\to\infty}[F(x)-\log\;x]$ を求めよ．
\[
 F(x)=\int_0^x \frac{t}{(t+1)(t+3)}\,dt\quad(x>0)
\]
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{\large 解答}\\
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\begin{eqnarray*}
F(x)
&=&\int_0^x \frac{t}{(t+1)(t+3)}\,dt\\
&=&\left[\dfrac{3\log \left(t+3\right)}{2}-\dfrac{\log \left(t+1\right)}{2}\right]_0^x\\
&=&\dfrac{3\log (x+3)}{2}-\dfrac{\log (x+1)}{2}-\dfrac{3\log (0+3)}{2}+\dfrac{\log (0+1)}{2}\\
&=&\dfrac{3\log (x+3)-\log (x+1)-3\log 3}{2}\\
\end{eqnarray*}

\begin{eqnarray*}
\lefteqn{\dlim_{x\to\infty}[F(x)-\log\;x]}\\
&=&\dlim_{x\to\infty}\left[\dfrac{3\log (x+3)-\log (x+1)-3\log 3}{2}-\log\;x\right]\\
&=&\dlim_{x\to\infty}\left[\dfrac{3\log (x+3)-\log (x+1)-2\log\;x-3\log 3}{2}\right]\\
&=&\dlim_{x\to\infty}\left[\dfrac{\log (x+3)^3/27x^2(x+1)}{2}\right]\\
&=&\dlim_{x\to\infty}\left[\dfrac{\log \{(x^3+9x^2+27x+27)/27x^2(x+1)\}}{2}\right]\\
\end{eqnarray*}
$\dfrac{x^3+9x^2+27x+27}{27x^2(x+1)}=\dfrac{x^3}{27x^2(x+1)}+\dfrac{9x^2}{27x^2(x+1)}+\dfrac{27x}{27x^2(x+1)}+\dfrac{27}{27x^2(x+1)}$
\begin{eqnarray*}
\dlim_{x\to\infty}\dfrac{x^3}{27x^2(x+1)}=\dfrac{1}{27}\\
\dlim_{x\to\infty}\dfrac{9x^2}{27x^2(x+1)}=0\\
\dlim_{x\to\infty}\dfrac{27x}{27x^2(x+1)}=0\\
\dlim_{x\to\infty}\dfrac{27}{27x^2(x+1)}=0
\end{eqnarray*}
より
$\dlim_{x\to\infty}\dfrac{x^3+9x^2+27x+27}{27x^2(x+1)}=\dfrac{1}{27}$
よって
\begin{eqnarray*}
\lefteqn{\dlim_{x\to\infty}[F(x)-\log\;x]}\\
&=&\dlim_{x\to\infty}\left[\dfrac{\log \{(x^3+9x^2+27x+27)/27x^2(x+1)\}}{2}\right]\\
&=&\dfrac{\log 1/27}{2}\\
&=&-\dfrac{3\log 3}{2}\\
\end{eqnarray*}
\\
\end{document}

