Задача Кузнецов Пределы 13-31

Условие задачи

Вычислить предел функции:

$\lim_{x\to \pi} \frac{\sin {\left(\frac{x^2}{\pi}\right)}}{2^{\sqrt{\sin {x}+1}}-2}$

Решение

Замена:

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

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

Получаем:

$\lim_{x\to \pi} \frac{\sin {\left(\frac{x^2}{\pi}\right)}}{2^{\sqrt{\sin {x}+1}}-2} = \lim_{y\to 0} \frac{\sin {\left(\frac{(y+\pi)^2}{\pi}\right)}}{2^{\sqrt{\sin {(y+\pi)}+1}}-2} =$

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

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

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

Воспользуемся заменой эквивалентных бесконечно малых:

$\sin {\left(y\cdot \frac{y+2\pi}{\pi}\right)} \sim y\cdot \frac{y+2\pi}{\pi}$, при $y \to 0 \left(y\cdot \frac{y+2\pi}{\pi}\to 0\right)$

$e^{\ln{2}\left(\sqrt{1-\sin {y}}-1\right)}-1 \sim \ln{2}\left(\sqrt{1-\sin {y}}-1\right)$, при $y \to 0 \left(\ln{2}\left(\sqrt{1-\sin {y}}-1\right)\to 0\right)$

Получаем:

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

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

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

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

Воспользуемся заменой эквивалентных бесконечно малых:

$\sin{y} \sim y$, при $y \to 0$

Получаем:

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

$= \frac{\frac{0+2\pi}{\pi}\left(\sqrt{1-\sin {0}}+1\right)}{2\ln{2}} = \frac{2\left(\sqrt{1}+1\right)}{2\ln{2}} = \frac{2}{\ln{2}}$