Задача 43188 ...
Условие
Решение
[m]\lim_{x \to\infty}(\frac{2x-1}{2x+3})^{1-4x}=\lim_{x \to\infty}(\frac{2x-1}{2x+3})^{-4x}\cdot(\frac{2x-1}{2x+3})^{1} =[/m]
Предел произведения равен произведению пределов:
[m]=\lim_{x \to\infty}(\frac{2x-1}{2x+3})^{-4x}\cdot \lim_{x \to\infty}(\frac{2x-1}{2x+3})^{1}[/m]
Так как
[m]\lim_{x \to\infty}(\frac{2x-1}{2x+3})^{1}=1^{1}=1[/m]
и
[m]\lim_{x \to\infty}(\frac{2x-1}{2x+3})^{-4x}=\lim_{x \to\infty}\frac{(\frac{2x-1}{2x})^{-4x}}{(\frac{2x+3}{2x})^{-4x}}=
=\lim_{x \to\infty}\frac{(1-\frac{1}{2x})^{-4x}}{(1+\frac{3}{2x})^{-4x}}=[/m]
[m]=\lim_{x \to\infty}\frac{((1-\frac{1}{2x})^{-2})^{2}}{((1+\frac{3}{2x})^{\frac{2x}{3}})^{-6}}=\frac{e^{2}}{e^{-6}}=e^{2-(-6)}=e^{8}[/m]