- 783 名前：１３２人目の素数さん [2008/01/02(水) 01:39:36 ]
- >>742
{\large $\displaystyle \lim_{x\rightarrow\infty}\frac{x^{m-n}[\{\log(1+x)\}^{n+1}-(\log x)^{n+1}]^{m}}{[\{\log(1+x)\}^{m+1}-(\log x)^{m+1}]^{n}}$}
$=\displaystyle \lim_{x\rightarrow\infty}\frac{x^{m-n}\{\log(1+x)-(\log x)\}^{m}[\{\log(1+x)\}^{n}+\{\log(1+x)\}^{n-1}(\log x)+\
text{・・・}+\{\log(1+x)\}^{n-k}(\log x)^
{\mathrm{k}}+\text{・・・}+(\log x)^{n}]^{m}}{\{\log(1+x)-(\log x)\}^{n}[\{\log(1+x)\}^{m}+\{\log(1+x)\}^{m-1}(\log x)+\text{・・・}+
\{\log(1+x)\}^{m-k}(\log x)^{\mathrm{k}}+\text{・・・}+(\log x)^{m}]^{n}}$
$=\displaystyle \lim_{x\rightarrow\infty}\frac{x^{m-n}\{\log(1+x)-(\log x)\}^{m-n}[\{\log(1+x)\}^{n}+\{\log(1+x)\}^{n-1}(\log x)+\text{・・・}+\
{\log(1+x)\}^{n-k}(\log x)^{\mathrm{k}}+\text{・・・}+(\log x)^
{n}]^{m}}{[\{\log(1+x)\}^{m}+\{\log(1+x)\}^{m-1}(\log x)+\text{・・・}+\{\log(1+x)\}^{m-k}(\log x)^{\mathrm{k}}+\text{・・・}+(\log x)^{m}]^{n}}$
=$\displaystyle \lim_{x\rightarrow\infty}\{x\log\frac{(1+x)}{x}\
}^{m-n}$・$\displaystyle \{\frac{[\{\log(1+x)\}^{n}+\{\log(1+x)\}^{n-1}(\log x)+\text{・・・}+
\{\log(1+x)\}^{n-k}(\log x)^{\mathrm{k}}+\text
{・・・}+(\log x)^{n}]^{\frac{1}{n}}}{[\{\log(1+x)\}^{m}+\
{\log(1+x)\}^{m-1}(\log x)+\text{・・・}+\{\log(1+x)\}^{m-k}(\log x)^
{\mathrm{k}}+\text{・・・}+(\log x)^{m}]^{\frac{1}{m}}}\}^{mn}$
|
 |
