Zadacha Kuznecov Predely 13-3

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Условие задачи

Вычислить предел функции:

<math>\lim_{x\to 2} \frac{\ln {\left(x-\sqrt[3]{2x-3}\right)}}{\sin {\left(\frac{\pi x}{2} \right)} - \sin {\left(\left(x-1 \right)\pi \right)}}</math>

Решение

Замена:

<math>x=y+2 \Rightarrow y=x -2</math>
<math>x\to 2 \Rightarrow y \to 0</math>

Получаем:

<math>\lim_{x\to 2} \frac{\ln {\left(x-\sqrt[3]{2x-3}\right)}}{\sin {\left(\frac{\pi x}{2} \right)} - \sin {\left(\left(x-1 \right)\pi \right)}} =\lim_{y\to 0} \frac{\ln {\left((y+2)-\sqrt[3]{2(y+2)-3}\right)}}{\sin {\left(\frac{\pi (y+2)}{2} \right)} - \sin {\left(\left((y+2)-1 \right)\pi \right)}} = </math>
<math>=\lim_{y\to 0} \frac{\ln {\left(y+2-\sqrt[3]{2y+1}\right)}}{\sin {\left(\frac{\pi y}{2}+\pi\right)} - \sin {\left(y \pi +\pi \right)}} = \lim_{y\to 0} \frac{\ln {\left(1+ \left(y+1-\sqrt[3]{2y+1}\right)\right)}}{\sin {y \pi} - \sin {\frac{\pi y}{2}}} = </math>
<math> = \lim_{y\to 0} \frac{\ln {\left(1+ \left(y+1-\sqrt[3]{2y+1}\right)\right)}}{2\sin {\frac{y \pi - \frac{\pi y}{2}}{2}}\cos{\frac{y \pi + \frac{\pi y}{2}}{2}}} =\lim_{y\to 0} \frac{\ln {\left(1+ \left(y+1-\sqrt[3]{2y+1}\right)\right)}}{2\sin {\frac{\pi y}{4}}\cos{\frac{3\pi y}{4}}} = </math>

Воспользуемся заменой эквивалентных бесконечно малых:

<math>\ln {\left(1+ \left(y+1-\sqrt[3]{2y+1}\right)\right)} \sim y+1-\sqrt[3]{2y+1}</math>, при <math>y \to 0\left(\left(y+1-\sqrt[3]{2y+1}\right) \to 0\right)</math>
<math>\sin {\frac{\pi y}{4}} \sim \frac{\pi y}{4}</math>, при <math>y \to 0\left(\frac{\pi y}{4}\to 0\right)</math>

Получаем:

<math> =\lim_{y\to 0} \frac{y+1-\sqrt[3]{2y+1}}{2\frac{\pi y}{4}\cos{\frac{3\pi y}{4}}} = </math>
<math> =\lim_{y\to 0} \frac{\left(y+1-\sqrt[3]{2y+1}\right)\left((y+1)^2+(y+1)\sqrt[3]{2y+1}+\sqrt[3]{(2y+1)^2}\right)}{\frac{\pi y}{2}\cos{\left(\frac{3\pi y}{4}\right)}\left((y+1)^2+(y+1)\sqrt[3]{2y+1}+\sqrt[3]{(2y+1)^2}\right)} = </math>
<math> =\lim_{y\to 0} \frac{(y+1)^3-(2y+1)}{\frac{\pi y}{2}\cos{\left(\frac{3\pi y}{4}\right)}\left((y+1)^2+(y+1)\sqrt[3]{2y+1}+\sqrt[3]{(2y+1)^2}\right)} = </math>
<math> =\lim_{y\to 0} \frac{y^3+3y^2+3y+1-2y-1}{\frac{\pi y}{2}\cos{\left(\frac{3\pi y}{4}\right)}\left((y+1)^2+(y+1)\sqrt[3]{2y+1}+\sqrt[3]{(2y+1)^2}\right)} = </math>
<math> =\lim_{y\to 0} \frac{y^3+3y^2+y}{\frac{\pi y}{2}\cos{\left(3\frac{\pi y}{4}\right)}\left((y+1)^2+(y+1)\sqrt[3]{2y+1}+\sqrt[3]{(2y+1)^2}\right)} = </math>
<math> =\lim_{y\to 0} \frac{y^2+3y+1}{\frac{\pi}{2}\cos{\left(3\frac{\pi y}{4}\right)}\left((y+1)^2+(y+1)\sqrt[3]{2y+1}+\sqrt[3]{(2y+1)^2}\right)} = </math>
<math> =\frac{0^2+3\cdot 0+1}{\frac{\pi}{2}\cos{\left(\frac{3\pi \cdot 0}{4}\right)}\left((0+1)^2+(0+1)\sqrt[3]{2\cdot 0+1}+\sqrt[3]{(2\cdot 0+1)^2}\right)} = </math>
<math> =\frac{1}{\frac{\pi}{2}\cdot 1\cdot \left(1+1\cdot \sqrt[3]{1}+\sqrt[3]{1^2}\right)} = \frac{1}{\frac{3\pi}{2}} = \frac{2}{3\pi}</math>