Exercice 14-10
Calculer la limite
\[
\lim_{\varepsilon\to
0^+}\log(1+\varepsilon)\int_1^\infty\frac{1}{x^{1+\varepsilon}}dx\,.
\]
Pour tout \(\varepsilon\gt 0\),
\[
\int_1^\infty\frac{dx}{x^{1+\varepsilon}}
=\lim_{L\to\infty}
\Bigl\{
\frac{x^{-\varepsilon}}{-\varepsilon}
\Bigr\}\Big|_1^L
=\frac{1}{\varepsilon}\lim_{L\to\infty}\Bigl(1-\frac{1}{L^\varepsilon}\Bigr)=\frac{1}{\varepsilon}\,.
\]
Ainsi, la limite demandée est
\[
\lim_{\varepsilon\to
0^+}\frac{\log(1+\varepsilon)}{\varepsilon}=1\,.
\]