Задача Кузнецов Пределы 13-29

Условие задачи

Вычислить предел функции:

$\lim_{x\to a\pi } \frac{\ln {\left(\cos{\left(\frac{x}{a}\right) +2}\right)}}{a^{a^{2}\pi ^{2}/x^{2} - a\pi/x}- a^{a\pi /x-1}}$

Решение

Замена:

$x=y+a\pi \Rightarrow y=x -a\pi$

$x\to a\pi \Rightarrow y \to 0$

Получаем:

$\lim_{x\to a\pi } \frac{\ln {\left(\cos{\left(\frac{x}{a}\right)} +2\right)}}{a^{a^{2}\pi ^{2}/x^{2} - a\pi/x}- a^{a\pi /x-1}} =$

$=\lim_{y\to 0} \frac{\ln {\left(\cos{\left(\frac{y+a\pi}{a}\right)} +2\right)}}{a^{a^{2}\pi ^{2}/(y+a\pi)^{2} - a\pi/(y+a\pi)}- a^{a\pi /(y+a\pi)-1}} =$

$=\lim_{y\to 0} \frac{\ln {\left(1+1-\cos{\left(\frac{y}{a}\right)}\right)}}{a^{a\pi /(y+a\pi)-1}\left(a^{a^{2}\pi^{2}/(y+a\pi)^{2} -2a\pi /(y+a\pi)+1}- 1\right)} =$

$=\lim_{y\to 0} \frac{\ln {\left(1+1-\cos{\left(\frac{y}{a}\right)}\right)}}{a^{a\pi /(y+a\pi)-1}\left(a^{\left(a\pi/(y+a\pi) - 1\right)^2}- 1\right)} =$

$=\lim_{y\to 0} \frac{\ln {\left(1+1-\cos{\left(\frac{y}{a}\right)}\right)}}{a^{a\pi /(y+a\pi)-1}\left(a^{\left(a\pi/(y+a\pi) - (y+a\pi)/(y+a\pi)\right)^2}- 1\right)} =$

$=\lim_{y\to 0} \frac{\ln {\left(1+1-\cos{\left(\frac{y}{a}\right)}\right)}}{a^{a\pi /(y+a\pi)-1}\left(\left(e^{\ln{a}}\right)^{\left(-y/(y+a\pi)\right)^2}- 1\right)} =$

$=\lim_{y\to 0} \frac{\ln {\left(1+1-\cos{\left(\frac{y}{a}\right)}\right)}}{a^{a\pi /(y+a\pi)-1}\left(e^{\ln{a}\cdot y^2 / (y+a\pi)^2}- 1\right)} =$

Воспользуемся заменой эквивалентных бесконечно малых:

$\ln {\left(1+1-\cos{\left(\frac{y}{a}\right)}\right)} \sim 1-\cos{\left(\frac{y}{a}\right)}$, при $y \to 0\left(1-\cos{\left(\frac{y}{a}\right)}\to 0\right)$

$e^{\ln{a}\cdot y^2 / (y+a\pi)^2}-1 \sim \ln{a}\cdot y^2 / (y+a\pi)^2$, при $y \to 0 \left(\ln{a}\cdot y^2 / (y+a\pi)^2\to 0\right)$

Получаем:

$=\lim_{y\to 0} \frac{1-\cos{\left(\frac{y}{a}\right)}}{a^{a\pi /(y+a\pi)-1}\left(\frac{\ln{a}\cdot y^2}{(y+a\pi)^2}\right)} =$

$=\lim_{y\to 0} \frac{\left(1-\cos{\left(\frac{y}{a}\right)}\right)(y+a\pi)^2}{a^{a\pi /(y+a\pi)-1}\left(\ln{a}\cdot y^2\right)} =$

Воспользуемся заменой эквивалентных бесконечно малых:

$1-\cos{\left(\frac{y}{a}\right)} \sim \frac{1}{2}\left(\frac{y}{a}\right)^2$, при $y \to 0 \left(\frac{y}{a}\to 0\right)$

Получаем:

$=\lim_{y\to 0} \frac{\frac{1}{2}\left(\frac{y}{a}\right)^2(y+a\pi)^2}{a^{a\pi /(y+a\pi)-1}\left(\ln{a}\cdot y^2\right)} =$

$=\lim_{y\to 0} \frac{(y+a\pi)^2}{2a^{2}\cdot a^{a\pi /(y+a\pi)-1}\left(\ln{a}\right)} =$

$=\frac{(0+a\pi)^2}{2a^{2}\cdot a^{a\pi /(0+a\pi)-1}\left(\ln{a}\right)} =$

$=\frac{a^2\cdot \pi^2}{2a^{2}\cdot a^{a\pi /a\pi-1}\left(\ln{a}\right)} =\frac{\pi^2}{2\cdot a^{0}\left(\ln{a}\right)} = \frac{\pi^2}{2\ln{a}}$