Задача №2015
Условие
Найти предел \(\lim_{x\to+\infty}\frac{\ln\left(2+e^{3x}\right)}{\ln\left(3+e^2x\right)}\).
Решение
\[
\lim_{x\to+\infty}\frac{\ln\left(2+e^{3x}\right)}{\ln\left(3+e^{2x}\right)}
=\lim_{x\to+\infty}\frac{\ln\left(e^{3x}\left(1+\frac{2}{e^{3x}}\right)\right)}{\ln\left(e^{2x}\left(1+\frac{3}{e^{2x}}\right)\right)}
=\lim_{x\to+\infty}\frac{3x+\ln\left(1+\frac{2}{e^{3x}}\right)}{2x+\ln\left(1+\frac{3}{e^{2x}}\right)}
=\lim_{x\to+\infty}\frac{3+\frac{\ln\left(1+\frac{2}{e^{3x}}\right)}{x}}{2+\frac{\ln\left(1+\frac{3}{e^{2x}}\right)}{x}}
=\frac{3}{2}.
\]
Ответ:
\(\frac{3}{2}\)