Zadacha Kuznecov Predely 14-5

Условие задачи

Вычислить предел функции:

$\lim _{x\to 0}{\frac {3^{2x}-5^{3x}}{\operatorname {arctg} {x}+x^{3}}}$

Решение

Первый способ

$\lim _{x\to 0}{\frac {3^{2x}-5^{3x}}{\operatorname {arctg} {x}+x^{3}}}=\lim _{x\to 0}{\frac {\left(9^{x}-1\right)-\left(125^{x}-1\right)}{\operatorname {arctg} {x}+x^{3}}}=$

$=\lim _{x\to 0}{\frac {\left(\left(e^{\ln {9}}\right)^{x}-1\right)-\left(\left(e^{\ln {125}}\right)^{x}-1\right)}{\operatorname {arctg} {x}+x^{3}}}=$

$=\lim _{x\to 0}{\frac {\left(e^{x\ln {9}}-1\right)-\left(e^{x\ln {125}}-1\right)}{\operatorname {arctg} {x}+x^{3}}}=$

$=\lim _{x\to 0}{\frac {{\frac {1}{x}}\left(\left(e^{x\ln {9}}-1\right)-\left(e^{x\ln {125}}-1\right)\right)}{{\frac {1}{x}}\left(\operatorname {arctg} {x}+x^{3}\right)}}=$

$={\frac {\lim \limits _{x\to 0}{\frac {1}{x}}\left(\left(e^{x\ln {9}}-1\right)-\left(e^{x\ln {125}}-1\right)\right)}{\lim \limits _{x\to 0}{\frac {1}{x}}\left(\operatorname {arctg} {x}+x^{3}\right)}}=$

$=\left(\lim _{x\to 0}{\frac {e^{x\ln {9}}-1}{x}}-\lim _{x\to 0}{\frac {e^{x\ln {125}}-1}{x}}\right)/\left(\lim _{x\to 0}{\frac {\operatorname {arctg} {x}}{x}}+\lim _{x\to 0}{\frac {x^{3}}{x}}\right)=$

Воспользуемся заменой эквивалентных бесконечно малых:

$e^{x\ln {9}}-1\sim x\ln {9}$, при $x\to 0(x\ln {9}\to 0)$

$e^{x\ln {125}}-1\sim x\ln {125}$, при $x\to 0(x\ln {125}\to 0)$

$\operatorname {arctg} {x}\sim x$, при $x\to 0$

Получаем:

$={\frac {\lim \limits _{x\to 0}{\frac {x\ln {9}}{x}}-\lim \limits _{x\to 0}{\frac {x\ln {125}}{x}}}{\lim \limits _{x\to 0}{\frac {x}{x}}+\lim \limits _{x\to 0}x^{2}}}={\frac {\lim \limits _{x\to 0}\ln {9}-\lim \limits _{x\to 0}\ln {125}}{\lim \limits _{x\to 0}1+0}}=$

$={\frac {\ln {9}-\ln {125}}{1+0}}=\ln {\frac {9}{125}}$

Второй способ

$\lim _{x\to 0}{\frac {3^{2x}-5^{3x}}{\operatorname {arctg} {x}+x^{3}}}=\lim _{x\to 0}{\frac {\left(3^{2x}-1\right)-\left(5^{3x}-1\right)}{\operatorname {arctg} {x}+x^{3}}}=$

Используем известные эквивалентности:

$3^{2x}-1\sim 2x\ln {3}$, при $x\to 0$

$5^{3x}-1\sim 3x\ln {5}$, при $x\to 0$

$\operatorname {arctg} {x}\sim x$, при $x\to 0$

$\lim _{x\to 0}{\frac {2x\ln {3}-3x\ln {5}}{x+x^{3}}}=\lim _{x\to 0}{\frac {2\ln {3}-3\ln {5}}{1+x^{2}}}=$

$=2\ln {3}-3\ln {5}=\ln {\frac {9}{125}}$