1040 lines
30 KiB
Markdown
1040 lines
30 KiB
Markdown
## 3 номер – Д 420
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#семестр_1 #высшая_математика
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### Пример:
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$\lim_{ x \to 1 } \dfrac{x^4-3x+2}{x^5-4x+3}$
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### Решение:
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$x = 1;$
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$x^4-3x+2 =0;$
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$x^5-4x+3=0;$
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$\text{Неопределённость } \frac{0}{0}$
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$(x - 1);$
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$x^{4}−3x+2=(x−1)(x^{3}+x^{2}+x+1)−3(x−1)=(x−1)(x^{3}+x^{2}+x−2);$
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$x^{5}−4x+3=(x−1)(x^{4}+x^{3}+x^{2}+x+1)−4(x−1)=(x−1)(x^{4}+x^{3}+x^{2}+x−3);$
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$x\neq1;$
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$\dfrac{(x−1)(x^{3}+x^{2}+x−2)}{(x−1)(x^{4}+x^{3}+x^{2}+x−3)}=\dfrac{x^{3}+x^{2}+x−2}{x^{4}+x^{3}+x^{2}+x−3}$
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$\lim_{ x \to 1 } \dfrac{x^{3}+x^{2}+x−2}{x^{4}+x^{3}+x^{2}+x−3} = \dfrac{1+1+1-2}{1+1+1+1-3}= \dfrac{1}{1}=1;$
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$\text{Ответ: }1$
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## 7 номер – Д 441
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### Пример:
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$\lim_{ x \to -2 } \dfrac{\sqrt[ 3 ]{ x-6 }+2}{x^{3}+8}$
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### Решение:
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$x=-2;$
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$\sqrt[3]{x-6}+2=0;$
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$x^3+8=0;$
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$\text{Неопределённость } \frac{0}{0}$
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$(x+2);$
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$(\sqrt[3]{x-6}+2)(\sqrt[3]{(x-6)^2}-2\sqrt[3]{x-6}+4)=x-6+8=x+2;$
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$x^3+8=(x+2)(x^2-2x+4);$
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$x\neq-2;$
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$\dfrac{\sqrt[3]{x-6}+2}{x^3+8}\cdot\dfrac{\sqrt[3]{(x-6)^2}-2\sqrt[3]{x-6}+4}{\sqrt[3]{(x6)^2}-2\sqrt[3]{x-6}+4}=\dfrac{x+2}{(x+2)(x^2-2x+4)(\sqrt[3]{(x-6)^2}-2\sqrt[3]{x-6}+4)}=$
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$=\dfrac{1}{(x^2-2x+4)(\sqrt[3]{(x-6)^2}-2\sqrt[3]{x-6}+4)}$
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$\lim_{ x \to -2 }\dfrac{1}{(x^2-2x+4)(\sqrt[3]{(x-6)^2}-2\sqrt[3]{x-6}+4)}=\dfrac{1}{(4+4+4)(4+4+4)}=\dfrac{1}{144};$
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$\text{Ответ: }\dfrac{1}{144}$
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## 11 номер – Д 446
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### Пример:
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$\lim_{ x \to 0 } \dfrac{\sqrt[ 3 ]{ 8+3x-x^{2} }-2}{x+x^{2}}$
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### Решение:
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$x=0;$
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$\sqrt[3]{8+3x-x^2}-2=\sqrt[3]{8}-2=0;$
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$x+x^2=0;$
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$\text{Неопределённость } \frac{0}{0}$
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$a=\sqrt[3]{8+3x-x^2}$
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$a-2=\dfrac{(a-2)(a^2+2a+4)}{a^2+2a+4}=\dfrac{a^3-8}{a^2+2a+4}$
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$a^3-8=(8+3x-x^2)-8=3x-x^2$
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$\dfrac{\sqrt[3]{8+3x-x^2}-2}{x+x^2}=\dfrac{3x-x^2}{(x+x^2)(a^2+2a+4)}$
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$x\neq0;$
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$3x-x^2=x(3-x);\ x+x^2=x(1+x)$
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$\dfrac{3x-x^2}{(x+x^2)(a^2+2a+4)}=\dfrac{x(3-x)}{x(1+x)(a^2+2a+4)}=\dfrac{3-x}{(1+x)(a^2+2a+4)}$
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$\lim_{x\to0}\dfrac{3-x}{(1+x)(a^2+2a+4)}=\dfrac{3}{1\cdot(2^2+2\cdot2+4)}=\dfrac{3}{12}=\dfrac{1}{4};$
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$\text{Ответ: }\dfrac{1}{4}$
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## 15 номер – Д 477
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### Пример:
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$\lim_{ x \to 0 } \dfrac{\cos x-\cos 3x}{x^{2}}$
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### Решение:
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$x=0;$
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$\cos x-\cos 3x=1-1=0;$
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$x^2=0;$
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$\text{Неопределённость } \frac{0}{0}$
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$\cos x-\cos 3x=-2\sin(\dfrac{x+3x}{2})\sin(\dfrac{x-3x}{2})=-2\sin(2x)\sin(-x)=2\sin(2x)\sin x;$
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$x\neq0;$
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$\dfrac{\cos x-\cos 3x}{x^2}=\dfrac{2\sin(2x)\sin x}{x^2}=4\cdot\dfrac{\sin(2x)}{2x}\cdot\dfrac{\sin x}{x};$
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$\lim_{ x \to 0 }4\cdot\dfrac{\sin(2x)}{2x}\cdot\dfrac{\sin x}{x}=4\cdot1\cdot1=4;$
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$\text{Ответ: }4$
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## 19 номер – Д 507
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### Пример:
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$\lim_{ x \to \infty } (\dfrac{x+2}{2x-1})^{x^{2}}$
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### Решение:
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$x\to\infty;$
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$\dfrac{x+2}{2x-1}\to\dfrac{1}{2};$
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$x>1;$
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$2x-1>0;$
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$\dfrac{x+2}{2x-1}>0;$
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$y=(\dfrac{x+2}{2x-1})^{x^2};$
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$\ln y=x^2\ln(\dfrac{x+2}{2x-1});$
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$\ln(\dfrac{x+2}{2x-1})\to\ln\dfrac{1}{2}<0;$
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$x^2\to+\infty;$
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$\ln y\to-\infty;$
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$y\to0;$
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$\text{Ответ: }0$
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## 22 номер – Д 510
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### Пример:
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$\lim_{ x \to \frac{\pi}{4}+0 } [\tan(\dfrac{\pi}{8}+x)]^{\tan2x}$
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### Решение:
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$x\to\dfrac{\pi}{4}+0;$
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$2x\to\dfrac{\pi}{2}+0;$
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$\tan2x\to-\infty;$
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$\dfrac{\pi}{8}+x\to\dfrac{3\pi}{8};$
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$\tan(\dfrac{\pi}{8}+x)\to\tan\dfrac{3\pi}{8}=\sqrt2+1>1;$
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$y=[\tan(\dfrac{\pi}{8}+x)]^{\tan2x};$
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$\ln y=\tan2x\cdot\ln(\tan(\dfrac{\pi}{8}+x));$
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$\ln(\tan(\dfrac{\pi}{8}+x))\to\ln(\sqrt2+1)>0;$
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$\tan2x\to-\infty;$
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$\ln y\to-\infty;$
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$y\to0;$
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$\text{Ответ: }0$
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## 23 номер – Д 511
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### Пример:
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$\lim_{ x \to \infty }(\dfrac{x^{2}-1}{x^2+1})^{\frac{x-1}{x+1}}$
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### Решение:
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$x\to\infty;$
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$\dfrac{x^{2}-1}{x^{2}+1}=\dfrac{1-\dfrac{1}{x^{2}}}{1+\dfrac{1}{x^{2}}}\to\dfrac{1-0}{1+0}=1;$
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$\dfrac{x-1}{x+1}\to1;$
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$\text{Вид }1^{1}$
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$y=(\dfrac{x^{2}-1}{x^{2}+1})^{\frac{x-1}{x+1}}$
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$\ln y=\dfrac{x-1}{x+1}\ln(\dfrac{x^{2}-1}{x^{2}+1})$
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$x>1;\ \dfrac{x^{2}-1}{x^{2}+1}>0$
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$\lim_{x\to\infty}\ln(\dfrac{x^{2}-1}{x^{2}+1})=\ln(\lim_{x\to\infty}\dfrac{x^{2}-1}{x^{2}+1})=\ln1=0;$
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$\lim_{x\to\infty}\dfrac{x-1}{x+1}=1;$
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$\lim_{x\to\infty}\ln y=1\cdot0=0$
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$\lim_{x\to\infty}y=\lim_{x\to\infty}e^{\ln y}=e^{0}=1;$
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$\text{Ответ: }1$
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## 26 номер – Д 514
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### Пример:
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$\lim_{ x \to 0 } \sqrt[ x ]{ 1-2x }$
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### Решение:
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$x=0;$
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$\sqrt[x]{1-2x}=(1-2x)^{\frac{1}{x}};$
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$1-2x=1;$
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$\text{Неопределённость }1^{\infty}$
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$x\to0;$
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$1-2x>0;$
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$y=(1-2x)^{\frac{1}{x}};$
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$\ln y=\dfrac{1}{x}\ln(1-2x);$
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$u=-2x;$
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$u\to0;$
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$\ln y=\dfrac{\ln(1+u)}{u}\cdot(-2);$
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$\lim_{ x \to 0 }\ln y=1\cdot(-2)=-2;$
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$\lim_{ x \to 0 }y=e^{-2};$
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$\text{Ответ: }e^{-2}$
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## 27 номер – Д 515
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### Пример:
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$\lim_{ x \to \infty } (\dfrac{x+a}{x-a})^{x}$
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### Решение:
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$x\to\infty;$
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$\dfrac{x+a}{x-a}\to1;$
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$\text{Неопределённость }1^{\infty}$
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$x\to\infty;$
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$x-a>0;$
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$\dfrac{x+a}{x-a}>0;$
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$y=(\dfrac{x+a}{x-a})^x;$
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$\ln y=x\ln(\dfrac{x+a}{x-a})=x\ln(1+\dfrac{2a}{x-a});$
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$u=\dfrac{2a}{x-a};$
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$u\to0;$
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$\ln y=\dfrac{\ln(1+u)}{u}\cdot x\cdot\dfrac{2a}{x-a};$
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$\lim_{ x \to \infty }\ln y=1\cdot 2a=2a;$
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$\lim_{ x \to \infty }y=e^{2a};$
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$\text{Ответ: }e^{2a}$
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## 30 номер – Д 519
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### Пример:
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$\lim_{ x \to 0 } (\dfrac{1+\tan x}{1+\sin x})^{ \frac{1}{\sin x} }$
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### Решение:
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a)
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$x=0;$
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$\dfrac{1+\tan x}{1+\sin x}=\dfrac{1+0}{1+0}=1;$
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$\text{Неопределённость }1^{\infty}$
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$y=(\dfrac{1+\tan x}{1+\sin x})^{\frac{1}{\sin x}};$
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$\ln y=\dfrac{1}{\sin x}\ln(\dfrac{1+\tan x}{1+\sin x});$
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$u=\dfrac{\tan x-\sin x}{1+\sin x};$
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$\dfrac{1+\tan x}{1+\sin x}=1+u;$
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$\ln y=\dfrac{\ln(1+u)}{u}\cdot\dfrac{u}{\sin x};$
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$\lim_{ x \to 0 }\dfrac{\ln(1+u)}{u}=1;$
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$\dfrac{u}{\sin x}=\dfrac{\tan x-\sin x}{\sin x(1+\sin x)}=\dfrac{\frac{\sin x}{\cos x}-\sin x}{\sin x(1+\sin x)}=\dfrac{\frac{1}{\cos x}-1}{1+\sin x}=\dfrac{1-\cos x}{\cos x(1+\sin x)};$
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$\lim_{ x \to 0 }\dfrac{u}{\sin x}=0;$
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$\lim_{ x \to 0 }\ln y=1\cdot0=0;$
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$\lim_{ x \to 0 }y=e^{0}=1;$
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$\text{Ответ: }1$
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## 34 номер – Д 523
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### Пример:
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$\lim_{ x \to \frac{\pi}{2} } (\sin x)^{ \tan x}$
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### Решение:
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$x\to\dfrac{\pi}{2};$
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$\sin x\to1;$
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$\tan x\to\pm\infty;$
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$\text{Неопределённость }1^{\infty}$
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$t=x-\dfrac{\pi}{2};$
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$t\to0;$
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$\sin x=\sin(\dfrac{\pi}{2}+t)=\cos t;$
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$\tan x=\tan(\dfrac{\pi}{2}+t)=-\cot t;$
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$\cos t>0\text{ при }t\to0;$
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$y=(\cos t)^{-\cot t};$
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$\ln y=-\cot t\cdot\ln(\cos t);$
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$u=\cos t-1;$
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$u\to0;$
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$\ln(\cos t)=\ln(1+u);$
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$\ln y=\dfrac{\ln(1+u)}{u}\cdot(-\cot t)\cdot(\cos t-1);$
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$\lim_{ t \to 0 }\dfrac{\ln(1+u)}{u}=1;$
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$(-\cot t)(\cos t-1)=-\dfrac{\cos t}{\sin t}(\cos t-1)=\dfrac{(1-\cos t)\cos t}{\sin t}=\dfrac{2\sin^2(\frac{t}{2})\cos t}{2\sin(\frac{t}{2})\cos(\frac{t}{2})}=\sin(\frac{t}{2})\cdot\dfrac{\cos t}{\cos(\frac{t}{2})};$
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$\lim_{ t \to 0 }(-\cot t)(\cos t-1)=0;$
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$\lim_{ x \to \frac{\pi}{2} }\ln y=1\cdot0=0;$
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$\lim_{ x \to \frac{\pi}{2} }y=e^{0}=1;$
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$\text{Ответ: }1$
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## 38 номер – Д 527
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### Пример:
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$\lim_{ n \to \infty } (\dfrac{n+1}{n-1})^{n}$
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### Решение:
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$n\to\infty;$
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$\dfrac{n+1}{n-1}\to1;$
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$n\to\infty;$
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$\text{Вид }1^{\infty}$
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$y=(\dfrac{n+1}{n-1})^{n}$
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$\ln y=n\ln(\dfrac{n+1}{n-1})=n\ln(1+\dfrac{2}{n-1})$
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$t=\dfrac{2}{n-1};\ t\to0+$
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$n=\dfrac{2}{t}+1$
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$\ln y=(\dfrac{2}{t}+1)\ln(1+t)=\dfrac{2}{t}\ln(1+t)+\ln(1+t)$
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$t>0;$
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$\dfrac{t}{1+t}\le\ln(1+t)\le t$
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$\dfrac{2}{1+t}\le\dfrac{2}{t}\ln(1+t)\le2$
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$t\to0+;$
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$\dfrac{2}{1+t}\to2;\ 2\to2$
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$\lim\limits_{t\to0+}\dfrac{2}{t}\ln(1+t)=2$
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$\lim\limits_{t\to0+}\ln(1+t)=\ln1=0$
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$\lim\limits_{n\to\infty}\ln y=2+0=2$
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$\lim\limits_{n\to\infty}y=e^{2}$
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$\text{Ответ: }e^{2}$
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## 42 номер – Д 531
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### Пример:
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$\lim_{ x \to a } \dfrac {\ln x - \ln a}{x-a}; (a>0)$
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### Решение:
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$x=a;$
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$\ln x-\ln a=0;$
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$x-a=0;$
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$\text{Неопределённость } \frac{0}{0}$
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$a>0;$
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$x>0;$
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$\ln x-\ln a=\ln\dfrac{x}{a}=\ln(1+\dfrac{x-a}{a});$
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$h=x-a;$
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$h\to0;$
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$y=\dfrac{\ln(1+\frac{h}{a})}{h};$
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$u=\dfrac{h}{a};$
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$u\to0;$
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$y=\dfrac{1}{a}\cdot\dfrac{\ln(1+u)}{u};$
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$\lim_{ x \to a }y=\dfrac{1}{a}\cdot1=\dfrac{1}{a};$
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$\text{Ответ: }\dfrac{1}{a}$
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## 46 номер – Д 476
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### Пример:
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$\lim_{ x \to 0 } \dfrac{\sin 5x - \sin 3x}{\sin x}$
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### Решение:
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$x=0;$
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$\sin5x-\sin3x=0;$
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$\sin x=0;$
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$\text{Неопределённость } \frac{0}{0}$
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$\sin5x-\sin3x=2\cos(\dfrac{5x+3x}{2})\sin(\dfrac{5x-3x}{2})=2\cos(4x)\sin x;$
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$x\neq0;$
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$\dfrac{\sin5x-\sin3x}{\sin x}=\dfrac{2\cos(4x)\sin x}{\sin x}=2\cos(4x);$
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$\lim_{ x \to 0 }2\cos(4x)=2\cos0=2;$
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$\text{Ответ: }2$
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## 50 номер – Д 460
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### Пример:
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$\lim_{ x \to 0+ } (\sqrt{ \dfrac{1}{x}+\sqrt{ \dfrac{1}{x}+\sqrt{ \dfrac{1}{x} } } }-\sqrt{ \dfrac{1}{x}-\sqrt{ \dfrac{1}{x}+\sqrt{ \dfrac{1}{x} } } })$
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### Решение:
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$x\to0+;$
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$\dfrac{1}{x}\to+\infty;$
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$t=\dfrac{1}{x};$
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$t\to+\infty;$
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$y=\sqrt{t+\sqrt{t+\sqrt t}}-\sqrt{t-\sqrt{t+\sqrt t}};$
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$A=t+\sqrt{t+\sqrt t};$
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$B=t-\sqrt{t+\sqrt t};$
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$y=\sqrt A-\sqrt B=\dfrac{A-B}{\sqrt A+\sqrt B}=\dfrac{2\sqrt{t+\sqrt t}}{\sqrt{t+\sqrt{t+\sqrt t}}+\sqrt{t-\sqrt{t+\sqrt t}}};$
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$\sqrt{t+\sqrt t}=\sqrt t\cdot\sqrt{1+\dfrac{1}{\sqrt t}};$
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$t\to+\infty;$
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$\sqrt{1+\dfrac{1}{\sqrt t}}\to1;$
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$\sqrt{t+\sqrt t}\sim\sqrt t;$
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$\dfrac{\sqrt{t+\sqrt t}}{t}=\dfrac{1}{\sqrt t}\cdot\sqrt{1+\dfrac{1}{\sqrt t}}\to0;$
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$\sqrt{t\pm\sqrt{t+\sqrt t}}=\sqrt t\cdot\sqrt{1\pm\dfrac{\sqrt{t+\sqrt t}}{t}}\sim\sqrt t;$
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$y\sim\dfrac{2\sqrt t}{\sqrt t+\sqrt t}=\dfrac{2\sqrt t}{2\sqrt t}=1;$
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$\lim_{ x \to 0+ }y=1;$
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$\text{Ответ: }1$
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## 53 номер – Д 496
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### Пример:
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$\lim_{ x \to \frac{\pi}{3} } \dfrac{\tan ^{3}x-3\tan x}{\cos(x+ \dfrac{\pi}{6})}$
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### Решение:
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$x=\dfrac{\pi}{3};$
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$\tan^3x-3\tan x=(\sqrt3)^3-3\sqrt3=0;$
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$\cos(x+\dfrac{\pi}{6})=\cos(\dfrac{\pi}{2})=0;$
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$\text{Неопределённость } \frac{0}{0}$
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$\tan^3x-3\tan x=\tan x(\tan^2x-3)=\tan x(\tan x-\sqrt3)(\tan x+\sqrt3);$
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$x\neq\dfrac{\pi}{3};$
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$\dfrac{\tan ^{3}x-3\tan x}{\cos(x+ \dfrac{\pi}{6})}=\dfrac{\tan x(\tan x-\sqrt3)(\tan x+\sqrt3)}{\cos(x+ \dfrac{\pi}{6})};$
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|
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$t=x-\dfrac{\pi}{3};\ t\to0;$
|
||
$\cos(x+\dfrac{\pi}{6})=\cos(t+\dfrac{\pi}{2})=-\sin t;$
|
||
$\tan x-\sqrt3=\tan x-\tan\dfrac{\pi}{3}=\dfrac{\sin(x-\dfrac{\pi}{3})}{\cos x\cos(\dfrac{\pi}{3})}=\dfrac{\sin t}{\cos x\cdot\dfrac12}=\dfrac{2\sin t}{\cos x};$
|
||
|
||
$x\neq\dfrac{\pi}{3};$
|
||
$\dfrac{\tan x(\tan x-\sqrt3)(\tan x+\sqrt3)}{\cos(x+ \dfrac{\pi}{6})}=\dfrac{\tan x\cdot\dfrac{2\sin t}{\cos x}\cdot(\tan x+\sqrt3)}{-\sin t}=-2\cdot\dfrac{\tan x(\tan x+\sqrt3)}{\cos x};$
|
||
|
||
$\lim_{ x \to \frac{\pi}{3} }-2\cdot\dfrac{\tan x(\tan x+\sqrt3)}{\cos x}=-2\cdot\dfrac{\sqrt3\cdot2\sqrt3}{\dfrac12}=-24;$
|
||
|
||
$\text{Ответ: }-24$
|
||
|
||
## 54 номер – Д 498
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ x \to \frac{\pi}{4} } \dfrac{1-\cot ^{3}x}{2-\cot x-\cot ^{3}x}$
|
||
|
||
### Решение:
|
||
|
||
$x=\dfrac{\pi}{4};$
|
||
$1-\cot^3x=1-1=0;$
|
||
$2-\cot x-\cot^3x=2-1-1=0;$
|
||
$\text{Неопределённость } \frac{0}{0}$
|
||
|
||
$u=\cot x;\ u\to1;$
|
||
$\dfrac{1-\cot ^{3}x}{2-\cot x-\cot ^{3}x}=\dfrac{1-u^3}{2-u-u^3};$
|
||
$1-u^3=(1-u)(1+u+u^2);$
|
||
$2-u-u^3=-(u^3+u-2)=-(u-1)(u^2+u+2)=(1-u)(u^2+u+2);$
|
||
|
||
$u\neq1;$
|
||
$\dfrac{1-u^3}{2-u-u^3}=\dfrac{(1-u)(1+u+u^2)}{(1-u)(u^2+u+2)}=\dfrac{1+u+u^2}{u^2+u+2};$
|
||
|
||
$\lim_{ x \to \frac{\pi}{4} }\dfrac{1+u+u^2}{u^2+u+2}=\dfrac{1+1+1}{1+1+2}=\dfrac{3}{4};$
|
||
|
||
$\text{Ответ: }\dfrac{3}{4}$
|
||
|
||
## 57 номер – Д 501
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ x \to 0 } \dfrac{\sqrt{ \cos x }-\sqrt[ 3 ]{ \cos x }}{\sin ^{2}x}$
|
||
|
||
### Решение:
|
||
|
||
$x=0;$
|
||
$\sqrt{\cos x}-\sqrt[3]{\cos x}=1-1=0;$
|
||
$\sin^2x=0;$
|
||
$\text{Неопределённость } \frac{0}{0}$
|
||
|
||
$u=\cos x-1;\ u\to0$
|
||
$\sqrt{\cos x}=(1+u)^{\frac12};\ \sqrt[3]{\cos x}=(1+u)^{\frac13}$
|
||
|
||
$(1+u)^{\alpha}-1\sim \alpha u\ (u\to0)$
|
||
$(1+u)^{\frac12}-1\sim\dfrac12u;\ (1+u)^{\frac13}-1\sim\dfrac13u$
|
||
|
||
$\sqrt{\cos x}-\sqrt[3]{\cos x}=\big((1+u)^{\frac12}-1\big)-\big((1+u)^{\frac13}-1\big)\sim(\dfrac12-\dfrac13)u=\dfrac16u=\dfrac16(\cos x-1)$
|
||
|
||
$\cos x-1\sim-\dfrac{x^2}{2};\ \sin x\sim x;\ \sin^2x\sim x^2$
|
||
|
||
$\dfrac{\sqrt{\cos x}-\sqrt[3]{\cos x}}{\sin^2x}\sim\dfrac{\dfrac16(\cos x-1)}{\sin^2x}\sim\dfrac{\dfrac16(-\dfrac{x^2}{2})}{x^2}=-\dfrac{1}{12}$
|
||
|
||
$\text{Ответ: }-\dfrac{1}{12}$
|
||
|
||
## 58 номер – Д 502
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ x \to 0 } \dfrac{\sqrt{ 1-\cos x^{2} }}{1-\cos x}$
|
||
|
||
### Решение:
|
||
|
||
$x=0;$
|
||
$\sqrt{1-\cos x^2}=\sqrt{0}=0;$
|
||
$1-\cos x=0;$
|
||
$\text{Неопределённость } \frac{0}{0}$
|
||
|
||
$1-\cos(x^2)\sim\dfrac{(x^2)^2}{2}=\dfrac{x^4}{2};$
|
||
$\sqrt{1-\cos x^2}\sim\sqrt{\dfrac{x^4}{2}}=\dfrac{x^2}{\sqrt2};$
|
||
$1-\cos x\sim\dfrac{x^2}{2};$
|
||
|
||
$\lim_{ x \to 0 }\dfrac{\sqrt{ 1-\cos x^{2} }}{1-\cos x}=\lim_{ x \to 0 }\dfrac{\dfrac{x^2}{\sqrt2}}{\dfrac{x^2}{2}}=\dfrac{2}{\sqrt2}=\sqrt2;$
|
||
|
||
$\text{Ответ: }\sqrt2$
|
||
|
||
## 61 номер – Д 505
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ x \to +\infty }(\sin \sqrt{ x+1 }-\sin \sqrt{ x })$
|
||
|
||
### Решение:
|
||
|
||
$\sin A-\sin B=2\cos(\dfrac{A+B}{2})\sin(\dfrac{A-B}{2});$
|
||
$A=\sqrt{x+1},\ B=\sqrt{x};$
|
||
$\sin \sqrt{ x+1 }-\sin \sqrt{ x }=2\cos(\dfrac{\sqrt{x+1}+\sqrt{x}}{2})\sin(\dfrac{\sqrt{x+1}-\sqrt{x}}{2});$
|
||
|
||
$\sqrt{x+1}-\sqrt{x}=\dfrac{(x+1)-x}{\sqrt{x+1}+\sqrt{x}}=\dfrac{1}{\sqrt{x+1}+\sqrt{x}};$
|
||
$\dfrac{\sqrt{x+1}-\sqrt{x}}{2}=\dfrac{1}{2(\sqrt{x+1}+\sqrt{x})}\to0;$
|
||
$|\cos(\dfrac{\sqrt{x+1}+\sqrt{x}}{2})|\le1;$
|
||
$\sin(\dfrac{\sqrt{x+1}-\sqrt{x}}{2})\sim\dfrac{\sqrt{x+1}-\sqrt{x}}{2};$
|
||
|
||
$|\sin \sqrt{ x+1 }-\sin \sqrt{ x }|\le2\cdot|\sin(\dfrac{\sqrt{x+1}-\sqrt{x}}{2})|\sim2\cdot\dfrac{\sqrt{x+1}-\sqrt{x}}{2}=\dfrac{1}{\sqrt{x+1}+\sqrt{x}}\to0;$
|
||
$\lim_{ x \to +\infty }(\sin \sqrt{ x+1 }-\sin \sqrt{ x })=0;$
|
||
|
||
$\text{Ответ: }0$
|
||
|
||
## 65 номер – Д 547
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ x \to 0 } \dfrac{e^{\alpha x}-e^{\beta x}}{\sin \alpha x - \sin \beta x}$
|
||
|
||
### Решение:
|
||
|
||
$x=0;$
|
||
$e^{\alpha x}-e^{\beta x}=1-1=0;$
|
||
$\sin\alpha x-\sin\beta x=0-0=0;$
|
||
$\text{Неопределённость } \frac{0}{0}$
|
||
|
||
$\alpha\neq\beta;$
|
||
$e^{\alpha x}-e^{\beta x}=e^{\beta x}(e^{(\alpha-\beta)x}-1);$
|
||
$\sin\alpha x-\sin\beta x=2\cos(\dfrac{(\alpha+\beta)x}{2})\sin(\dfrac{(\alpha-\beta)x}{2});$
|
||
|
||
$t=\dfrac{(\alpha-\beta)x}{2};\ t\to0;$
|
||
$\dfrac{e^{\alpha x}-e^{\beta x}}{\sin \alpha x - \sin \beta x}=\dfrac{e^{\beta x}(e^{2t}-1)}{2\cos(\dfrac{(\alpha+\beta)x}{2})\sin t}=\dfrac{e^{\beta x}}{\cos(\dfrac{(\alpha+\beta)x}{2})}\cdot\dfrac{e^{2t}-1}{2t}\cdot\dfrac{t}{\sin t};$
|
||
|
||
$e^{\beta x}\to1;$
|
||
$\cos(\dfrac{(\alpha+\beta)x}{2})\to1;$
|
||
$\dfrac{e^{2t}-1}{2t}\to1;$
|
||
$\dfrac{t}{\sin t}\to1;$
|
||
$\lim_{ x \to 0 }\dfrac{e^{\alpha x}-e^{\beta x}}{\sin \alpha x - \sin \beta x}=1;$
|
||
|
||
$\text{Ответ: }1$
|
||
|
||
## 69 номер – Д 574
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ x \to 1 } (2-x)^{\sec \frac{\pi x}{2}}$
|
||
|
||
### Решение:
|
||
|
||
$x\to1;$
|
||
$2-x\to1;$
|
||
$\sec\dfrac{\pi x}{2}=\dfrac{1}{\cos(\dfrac{\pi x}{2})}\to\infty;$
|
||
$\text{Неопределённость }1^{\infty}$
|
||
|
||
$t=x-1;\ t\to0;$
|
||
$y=(1-t)^{\sec(\dfrac{\pi}{2}+\dfrac{\pi t}{2})};$
|
||
$\ln y=\sec(\dfrac{\pi}{2}+\dfrac{\pi t}{2})\ln(1-t);$
|
||
|
||
$\cos(\dfrac{\pi}{2}+\dfrac{\pi t}{2})=-\sin(\dfrac{\pi t}{2});$
|
||
$\sec(\dfrac{\pi}{2}+\dfrac{\pi t}{2})=-\dfrac{1}{\sin(\dfrac{\pi t}{2})};$
|
||
$\ln y=-\dfrac{\ln(1-t)}{\sin(\dfrac{\pi t}{2})}=(-\dfrac{\ln(1-t)}{t})\cdot\dfrac{t}{\sin(\dfrac{\pi t}{2})};$
|
||
|
||
$-\dfrac{\ln(1-t)}{t}\to1;$
|
||
$\dfrac{t}{\sin(\dfrac{\pi t}{2})}=\dfrac{2}{\pi}\cdot\dfrac{\frac{\pi t}{2}}{\sin(\frac{\pi t}{2})}\to\dfrac{2}{\pi};$
|
||
$\lim_{ x \to 1 }\ln y=\dfrac{2}{\pi};$
|
||
$\lim_{ x \to 1 }y=e^{2/\pi};$
|
||
|
||
$\text{Ответ: }e^{2/\pi}$
|
||
|
||
## 73 номер – Д 577
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ x \to +\infty } \dfrac{sh\sqrt{ x^{2}+x }-sh\sqrt{ x^{2}-x }}{ch(x)}$
|
||
|
||
$ch(x)=\dfrac{1}{2}(e^{x}+e^{-x})$
|
||
$sh(x)=\dfrac{1}{2}(e^{x}-e^{-x})$
|
||
$th(x)=\dfrac{sh(x)}{ch(x)}$
|
||
|
||
### Решение:
|
||
|
||
$x\to+\infty;$
|
||
$sh\sqrt{x^2+x}-sh\sqrt{x^2-x}\to\infty;$
|
||
$chx\to\infty;$
|
||
$\text{Неопределённость } \frac{\infty}{\infty}$
|
||
|
||
$A=\sqrt{x^2+x};\ B=\sqrt{x^2-x};$
|
||
$\dfrac{shA-shB}{chx}=\dfrac{\frac12(e^A-e^{-A})-\frac12(e^B-e^{-B})}{\frac12(e^x+e^{-x})}=\dfrac{e^A-e^B+e^{-B}-e^{-A}}{e^x+e^{-x}};$
|
||
|
||
$\dfrac{shA-shB}{chx}=\dfrac{e^{A-x}-e^{B-x}+e^{-(B+x)}-e^{-(A+x)}}{1+e^{-2x}};$
|
||
$x\to+\infty;$
|
||
$e^{-(B+x)}\to0;$
|
||
$e^{-(A+x)}\to0;$
|
||
$1+e^{-2x}\to1;$
|
||
$\lim_{ x \to +\infty }\dfrac{shA-shB}{chx}=\lim_{ x \to +\infty }(e^{A-x}-e^{B-x});$
|
||
|
||
$A-x=\sqrt{x^2+x}-x=\dfrac{x}{\sqrt{x^2+x}+x}=\dfrac{1}{\sqrt{1+\frac{1}{x}}+1}\to\dfrac12;$
|
||
$B-x=\sqrt{x^2-x}-x=\dfrac{-x}{\sqrt{x^2-x}+x}=-\dfrac{1}{\sqrt{1-\frac{1}{x}}+1}\to-\dfrac12;$
|
||
|
||
$e^{A-x}\to e^{1/2};$
|
||
$e^{B-x}\to e^{-1/2};$
|
||
$\lim_{ x \to +\infty }\dfrac{sh\sqrt{ x^{2}+x }-sh\sqrt{ x^{2}-x }}{ch(x)}=e^{1/2}-e^{-1/2}=2sh(\dfrac12);$
|
||
|
||
$\text{Ответ: }e^{1/2}-e^{-1/2}$
|
||
|
||
## 77 номер – Д 581
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ x \to \infty } \arcsin \dfrac{1-x}{1+x}$
|
||
|
||
### Решение:
|
||
|
||
$\dfrac{1-x}{1+x}=-\dfrac{x-1}{x+1}=-(1-\dfrac{2}{x+1})=-1+\dfrac{2}{x+1};$
|
||
$x\to\infty;$
|
||
$\dfrac{1-x}{1+x}\to-1;$
|
||
$\arcsin$ непрерывна на $[-1;1];$
|
||
$\lim_{ x \to \infty }\arcsin \dfrac{1-x}{1+x}=\arcsin(-1)=-\dfrac{\pi}{2};$
|
||
|
||
$\text{Ответ: }-\dfrac{\pi}{2}$
|
||
|
||
## 80 номер – Д 584
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ x \to -\infty } \text{arccot} \dfrac{x}{\sqrt{ 1+x^{2} }}$
|
||
|
||
### Решение:
|
||
|
||
$x\to-\infty;$
|
||
$\sqrt{1+x^2}=|x|\sqrt{1+\dfrac{1}{x^2}}=-x\sqrt{1+\dfrac{1}{x^2}};$
|
||
$\dfrac{x}{\sqrt{1+x^2}}=\dfrac{x}{-x\sqrt{1+\frac{1}{x^2}}}=-\dfrac{1}{\sqrt{1+\frac{1}{x^2}}}\to-1;$
|
||
|
||
$\text{arccot}$ берём со значениями в $(0;\pi);$
|
||
$\lim_{ x \to -\infty }\text{arccot}\dfrac{x}{\sqrt{ 1+x^{2} }}=\text{arccot}(-1)=\dfrac{3\pi}{4};$
|
||
|
||
$\text{Ответ: }\dfrac{3\pi}{4}$
|
||
|
||
## 81 номер – Д 585
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ h \to 0 } \dfrac{\text{arccot}(x+h)-\text{arccot}x}{h}$
|
||
|
||
### Решение:
|
||
|
||
$x=\text{const};$
|
||
$\text{arccot}\,t=\dfrac{\pi}{2}-\arctan t;$
|
||
$\dfrac{\text{arccot}(x+h)-\text{arccot}x}{h}=\dfrac{(\dfrac{\pi}{2}-\arctan(x+h))-(\dfrac{\pi}{2}-\arctan x)}{h}=-\dfrac{\arctan(x+h)-\arctan x}{h};$
|
||
|
||
$y=\arctan(x+h),\ z=\arctan x;$
|
||
$\tan(y-z)=\dfrac{\tan y-\tan z}{1+\tan y\tan z}=\dfrac{(x+h)-x}{1+x(x+h)}=\dfrac{h}{1+x^{2}+xh};$
|
||
$y-z=\arctan(\dfrac{h}{1+x^{2}+xh});$
|
||
|
||
$u=\dfrac{h}{1+x^{2}+xh};\ u\to0;$
|
||
$\dfrac{\arctan(x+h)-\arctan x}{h}=\dfrac{\arctan u}{u}\cdot\dfrac{u}{h};$
|
||
|
||
$t=\arctan u;\ u=\tan t;$
|
||
$\dfrac{\arctan u}{u}=\dfrac{t}{\tan t}=\dfrac{t}{\sin t}\cdot\cos t\to1;$
|
||
$\dfrac{u}{h}=\dfrac{1}{1+x^{2}+xh}\to\dfrac{1}{1+x^{2}};$
|
||
$\lim_{ h \to 0 }\dfrac{\arctan(x+h)-\arctan x}{h}=\dfrac{1}{1+x^{2}};$
|
||
$\lim_{ h \to 0 }\dfrac{\text{arccot}(x+h)-\text{arccot}x}{h}=-\dfrac{1}{1+x^{2}};$
|
||
|
||
$\text{Ответ: }-\dfrac{1}{1+x^{2}}$
|
||
|
||
## 84 номер – Д 590
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ n \to \infty }[1+\dfrac{(-1)^{n}}{n}]^{\text{cosec}(\pi \sqrt{ 1+n^{2} })}$
|
||
|
||
### Решение:
|
||
|
||
$a_n=(1+\dfrac{(-1)^n}{n})^{\text{cosec}(\pi\sqrt{1+n^2})};$
|
||
$n\ge2;$
|
||
$1+\dfrac{(-1)^n}{n}>0;$
|
||
$\sin(\pi\sqrt{1+n^2})\neq0;$
|
||
$\text{Неопределённость }1^{\infty}$
|
||
|
||
$\ln a_n=\text{cosec}(\pi\sqrt{1+n^2})\cdot\ln(1+\dfrac{(-1)^n}{n});$
|
||
|
||
$\delta_n=\sqrt{1+n^2}-n=\dfrac{(1+n^2)-n^2}{\sqrt{1+n^2}+n}=\dfrac{1}{\sqrt{1+n^2}+n}\sim\dfrac{1}{2n};$
|
||
$\sin(\pi\sqrt{1+n^2})=\sin(\pi(n+\delta_n))=\sin(\pi n+\pi\delta_n)=(-1)^n\sin(\pi\delta_n);$
|
||
$\sin(\pi\delta_n)\sim\pi\delta_n;$
|
||
$\sin(\pi\sqrt{1+n^2})\sim(-1)^n\pi\cdot\dfrac{1}{2n}=(-1)^n\dfrac{\pi}{2n};$
|
||
$\text{cosec}(\pi\sqrt{1+n^2})=\dfrac{1}{\sin(\pi\sqrt{1+n^2})}\sim(-1)^n\dfrac{2n}{\pi};$
|
||
|
||
$\ln(1+\dfrac{(-1)^n}{n})\sim\dfrac{(-1)^n}{n};$
|
||
$\ln a_n\sim((-1)^n\dfrac{2n}{\pi})\cdot(\dfrac{(-1)^n}{n})=\dfrac{2}{\pi};$
|
||
$\lim_{n\to\infty}\ln a_n=\dfrac{2}{\pi};$
|
||
$\lim_{n\to\infty}a_n=e^{2/\pi};$
|
||
|
||
$\text{Ответ: }e^{2/\pi}$
|
||
|
||
## 85 номер – Д 591
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ x \to 0 } \dfrac{1}{x^{100}}e^{-\frac{1}{x^{2}}}$
|
||
|
||
### Решение:
|
||
|
||
$x\neq0;$
|
||
$\dfrac{1}{x^{100}}e^{-1/x^{2}}=\dfrac{1}{(x^2)^{50}}e^{-1/x^{2}};$
|
||
$t=\dfrac{1}{x^2};\ t\to+\infty;$
|
||
$\dfrac{1}{(x^2)^{50}}e^{-1/x^{2}}=t^{50}e^{-t};$
|
||
|
||
$e^{t}=1+t+\dfrac{t^2}{2!}+\dots+\dfrac{t^{51}}{51!}+\dots\ge\dfrac{t^{51}}{51!};\ (t>0);$
|
||
$e^{-t}\le\dfrac{51!}{t^{51}};$
|
||
$0\le t^{50}e^{-t}\le t^{50}\cdot\dfrac{51!}{t^{51}}=\dfrac{51!}{t}\to0;$
|
||
$\lim_{ x \to 0 }\dfrac{1}{x^{100}}e^{-1/x^{2}}=0;$
|
||
|
||
$\text{Ответ: }0$
|
||
|
||
## 88 номер – Д 593Б
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ x \to +\infty } (\sqrt{ x^{2}+x }-x)$
|
||
|
||
### Решение:
|
||
|
||
$\sqrt{x^2+x}-x=\dfrac{(\sqrt{x^2+x}-x)(\sqrt{x^2+x}+x)}{\sqrt{x^2+x}+x}=\dfrac{x^2+x-x^2}{\sqrt{x^2+x}+x}=\dfrac{x}{\sqrt{x^2+x}+x};$
|
||
$x>0;$
|
||
$\dfrac{x}{\sqrt{x^2+x}+x}=\dfrac{x}{x\sqrt{1+\frac{1}{x}}+x}=\dfrac{1}{\sqrt{1+\frac{1}{x}}+1};$
|
||
$x\to+\infty;$
|
||
$\dfrac{1}{\sqrt{1+\frac{1}{x}}+1}\to\dfrac{1}{1+1}=\dfrac{1}{2};$
|
||
|
||
$\text{Ответ: }\dfrac{1}{2}$
|
||
|
||
## 89 номер – Д 594А
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ x \to -\infty }(\sqrt{ 1+x+x^{2} }-\sqrt{ 1-x+x^{2} })$
|
||
|
||
### Решение:
|
||
|
||
$\sqrt{x^2+x+1}-\sqrt{x^2-x+1}=\dfrac{(x^2+x+1)-(x^2-x+1)}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}=\dfrac{2x}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}};$
|
||
$x\to-\infty;$
|
||
$\sqrt{x^2+x+1}=|x|\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}}=-x\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}};$
|
||
$\sqrt{x^2-x+1}=|x|\sqrt{1-\dfrac{1}{x}+\dfrac{1}{x^2}}=-x\sqrt{1-\dfrac{1}{x}+\dfrac{1}{x^2}};$
|
||
|
||
$\dfrac{2x}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}=\dfrac{2x}{-x(\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}}+\sqrt{1-\dfrac{1}{x}+\dfrac{1}{x^2}})}=-\dfrac{2}{\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}}+\sqrt{1-\dfrac{1}{x}+\dfrac{1}{x^2}}};$
|
||
$x\to-\infty;$
|
||
$-\dfrac{2}{1+1}=-1;$
|
||
|
||
$\text{Ответ: }-1$
|
||
|
||
## 92 номер – Д 1319
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ x \to 0 } \dfrac{ch(x)-\cos x}{x^{2}}$
|
||
|
||
### Решение:
|
||
|
||
$x=0;$
|
||
$ch(x)-\cos x=1-1=0;$
|
||
$x^2=0;$
|
||
$\text{Неопределённость } \frac{0}{0}$
|
||
|
||
$\lim_{ x \to 0 }\dfrac{ch(x)-\cos x}{x^{2}}=\lim_{ x \to 0 }\dfrac{(ch(x)-\cos x)'}{(x^{2})'}=\lim_{ x \to 0 }\dfrac{sh(x)+\sin x}{2x};$
|
||
$x=0;$
|
||
$sh(x)+\sin x=0+0=0;$
|
||
$2x=0;$
|
||
$\text{Неопределённость } \frac{0}{0}$
|
||
|
||
$\lim_{ x \to 0 }\dfrac{sh(x)+\sin x}{2x}=\lim_{ x \to 0 }\dfrac{(sh(x)+\sin x)'}{(2x)'}=\lim_{ x \to 0 }\dfrac{ch(x)+\cos x}{2}=\dfrac{1+1}{2}=1;$
|
||
|
||
$\text{Ответ: }1$
|
||
|
||
## 96 номер – Д 1323
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ x \to 0 } \dfrac{x \cot x -1}{x^{2}}$
|
||
|
||
### Решение:
|
||
|
||
$x=0;$
|
||
$x\cot x-1=0;$
|
||
$x^2=0;$
|
||
$\text{Неопределённость } \frac{0}{0}$
|
||
|
||
$x\neq0;$
|
||
$x\cot x-1=x\dfrac{\cos x}{\sin x}-1=\dfrac{x\cos x-\sin x}{\sin x};$
|
||
$\dfrac{x\cot x-1}{x^2}=\dfrac{x\cos x-\sin x}{x^2\sin x};$
|
||
|
||
$\lim_{ x \to 0 }\dfrac{x\cos x-\sin x}{x^2\sin x}=\lim_{ x \to 0 }\dfrac{(x\cos x-\sin x)'}{(x^2\sin x)'}=\lim_{ x \to 0 }\dfrac{-x\sin x}{2x\sin x+x^2\cos x};$
|
||
$x\neq0;$
|
||
$\dfrac{-x\sin x}{2x\sin x+x^2\cos x}=\dfrac{-\sin x}{2\sin x+x\cos x};$
|
||
$x=0;$
|
||
$-\sin x=0;$
|
||
$2\sin x+x\cos x=0;$
|
||
$\text{Неопределённость } \frac{0}{0}$
|
||
|
||
$\lim_{ x \to 0 }\dfrac{-\sin x}{2\sin x+x\cos x}=\lim_{ x \to 0 }\dfrac{(-\sin x)'}{(2\sin x+x\cos x)'}=\lim_{ x \to 0 }\dfrac{-\cos x}{2\cos x+\cos x-x\sin x}=\dfrac{-1}{3}=-\dfrac{1}{3};$
|
||
|
||
$\text{Ответ: }-\dfrac{1}{3}$
|
||
|
||
## 100 номер – Д 1327
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ x \to 0 } \dfrac{\arcsin 2x -2\arcsin x}{x^{3}}$
|
||
|
||
### Решение:
|
||
|
||
$x=0;$
|
||
$\arcsin2x-2\arcsin x=0;$
|
||
$x^3=0;$
|
||
$\text{Неопределённость } \frac{0}{0}$
|
||
|
||
$\lim_{ x \to 0 }\dfrac{\arcsin 2x -2\arcsin x}{x^{3}}=\lim_{ x \to 0 }\dfrac{(\arcsin 2x -2\arcsin x)'}{(x^{3})'}=\lim_{ x \to 0 }\dfrac{\dfrac{2}{\sqrt{1-4x^2}}-\dfrac{2}{\sqrt{1-x^2}}}{3x^2};$
|
||
$x=0;$
|
||
$\dfrac{2}{\sqrt{1-4x^2}}-\dfrac{2}{\sqrt{1-x^2}}=0;$
|
||
$3x^2=0;$
|
||
$\text{Неопределённость } \frac{0}{0}$
|
||
|
||
$\lim_{ x \to 0 }\dfrac{\dfrac{2}{\sqrt{1-4x^2}}-\dfrac{2}{\sqrt{1-x^2}}}{3x^2}=\lim_{ x \to 0 }\dfrac{(\dfrac{2}{\sqrt{1-4x^2}}-\dfrac{2}{\sqrt{1-x^2}})'}{(3x^2)'}=\lim_{ x \to 0 }\dfrac{\dfrac{8x}{(1-4x^2)^{3/2}}-\dfrac{2x}{(1-x^2)^{3/2}}}{6x};$
|
||
$x=0;$
|
||
$\dfrac{8x}{(1-4x^2)^{3/2}}-\dfrac{2x}{(1-x^2)^{3/2}}=0;$
|
||
$6x=0;$
|
||
$\text{Неопределённость } \frac{0}{0}$
|
||
|
||
$\lim_{ x \to 0 }\dfrac{\dfrac{8x}{(1-4x^2)^{3/2}}-\dfrac{2x}{(1-x^2)^{3/2}}}{6x}=\lim_{ x \to 0 }\dfrac{(\dfrac{8x}{(1-4x^2)^{3/2}}-\dfrac{2x}{(1-x^2)^{3/2}})'}{(6x)'}=$
|
||
$=\lim_{ x \to 0 }\dfrac{\dfrac{96x^2}{(1-4x^2)^{5/2}}-\dfrac{6x^2}{(1-x^2)^{5/2}}+\dfrac{8}{(1-4x^2)^{3/2}}-\dfrac{2}{(1-x^2)^{3/2}}}{6}=\dfrac{8-2}{6}=1;$
|
||
|
||
$\text{Ответ: }1$
|
||
|
||
## 104 номер – Д 1331
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ x \to 0 } \dfrac{\ln(\sin ax)}{\ln(\sin bx)}$
|
||
|
||
### Решение:
|
||
|
||
$x\to0+;$
|
||
$\ln(\sin ax)\to-\infty;$
|
||
$\ln(\sin bx)\to-\infty;$
|
||
$\text{Неопределённость }\dfrac{-\infty}{-\infty}$
|
||
|
||
$L=\lim\limits_{x\to0+}\dfrac{\ln(\sin ax)}{\ln(\sin bx)}$
|
||
$L=\lim\limits_{x\to0+}\dfrac{\big(\ln(\sin ax)\big)'}{\big(\ln(\sin bx)\big)'}=\lim\limits_{x\to0+}\dfrac{a\cot(ax)}{b\cot(bx)}=\dfrac{a}{b}\lim\limits_{x\to0+}\dfrac{\cot(ax)}{\cot(bx)}$
|
||
|
||
$\cot t=\dfrac{\cos t}{\sin t}$
|
||
$\dfrac{\cot(ax)}{\cot(bx)}=\dfrac{\cos(ax)\sin(bx)}{\sin(ax)\cos(bx)}$
|
||
|
||
$\cos(ax)\sin(bx)\to1\cdot0=0;$
|
||
$\sin(ax)\cos(bx)\to0\cdot1=0;$
|
||
$\text{Неопределённость }\dfrac{0}{0}$
|
||
|
||
$\lim\limits_{x\to0+}\dfrac{\cos(ax)\sin(bx)}{\sin(ax)\cos(bx)}=\lim\limits_{x\to0+}\dfrac{\big(\cos(ax)\sin(bx)\big)'}{\big(\sin(ax)\cos(bx)\big)'}$
|
||
|
||
$\big(\cos(ax)\sin(bx)\big)'=-a\sin(ax)\sin(bx)+b\cos(ax)\cos(bx)$
|
||
$\big(\sin(ax)\cos(bx)\big)'=a\cos(ax)\cos(bx)-b\sin(ax)\sin(bx)$
|
||
|
||
$\lim\limits_{x\to0+}\dfrac{-a\sin(ax)\sin(bx)+b\cos(ax)\cos(bx)}{a\cos(ax)\cos(bx)-b\sin(ax)\sin(bx)}=\dfrac{b}{a}$
|
||
|
||
$L=\dfrac{a}{b}\cdot\dfrac{b}{a}=1$
|
||
|
||
$\text{Ответ: }1$
|
||
|
||
## 108 номер – Д 1335
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ x \to 0 } \dfrac{Arsh(sh(x))-Arsh(\sin x)}{sh(x) - \sin x}$
|
||
|
||
$Arsh(x)=\ln(x+\sqrt{ 1+x^{2} })$
|
||
|
||
### Решение:
|
||
|
||
$x=0;$
|
||
$Arsh(shx)-Arsh(\sin x)=0;$
|
||
$shx-\sin x=0;$
|
||
$\text{Неопределённость } \frac{0}{0}$
|
||
|
||
$\lim_{ x \to 0 }\dfrac{Arsh(shx)-Arsh(\sin x)}{shx-\sin x}=\lim_{ x \to 0 }\dfrac{(Arsh(shx)-Arsh(\sin x))'}{(shx-\sin x)'}=$
|
||
$=\lim_{ x \to 0 }\dfrac{\dfrac{chx}{\sqrt{1+sh^2x}}-\dfrac{\cos x}{\sqrt{1+\sin^2x}}}{chx-\cos x};$
|
||
$x=0;$
|
||
$\dfrac{chx}{\sqrt{1+sh^2x}}-\dfrac{\cos x}{\sqrt{1+\sin^2x}}=0;$
|
||
$chx-\cos x=0;$
|
||
$\text{Неопределённость } \frac{0}{0}$
|
||
|
||
$1+sh^2x=ch^2x;$
|
||
$\sqrt{1+sh^2x}=chx;$
|
||
$\lim_{ x \to 0 }\dfrac{\dfrac{chx}{\sqrt{1+sh^2x}}-\dfrac{\cos x}{\sqrt{1+\sin^2x}}}{chx-\cos x}=\lim_{ x \to 0 }\dfrac{1-\dfrac{\cos x}{\sqrt{1+\sin^2x}}}{chx-\cos x};$
|
||
|
||
$\lim_{ x \to 0 }\dfrac{1-\dfrac{\cos x}{\sqrt{1+\sin^2x}}}{chx-\cos x}=\lim_{ x \to 0 }\dfrac{(1-\dfrac{\cos x}{\sqrt{1+\sin^2x}})'}{(chx-\cos x)'}=$
|
||
$=\lim_{ x \to 0 }\dfrac{\dfrac{2\sin x}{(1+\sin^2x)^{3/2}}}{shx+\sin x};$
|
||
$x=0;$
|
||
$\dfrac{2\sin x}{(1+\sin^2x)^{3/2}}=0;$
|
||
$shx+\sin x=0;$
|
||
$\text{Неопределённость } \frac{0}{0}$
|
||
|
||
$\lim_{ x \to 0 }\dfrac{\dfrac{2\sin x}{(1+\sin^2x)^{3/2}}}{shx+\sin x}=\lim_{ x \to 0 }\dfrac{(\dfrac{2\sin x}{(1+\sin^2x)^{3/2}})'}{(shx+\sin x)'}=$
|
||
$=\lim_{ x \to 0 }\dfrac{\dfrac{2\cos x\cos2x}{(1+\sin^2x)^{5/2}}}{chx+\cos x}=\dfrac{2}{2}=1;$
|
||
|
||
$\text{Ответ: }1$
|
||
|
||
## 111 номер – Д 1338
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ x \to 0 } \dfrac{e^{- \frac{1}{x^{2}}}}{x^{100}}$
|
||
|
||
### Решение:
|
||
|
||
$x\to0;$
|
||
$e^{-1/x^2}\to0;$
|
||
$x^{100}\to0;$
|
||
$\text{Неопределённость } \frac{0}{0}$
|
||
|
||
$x\neq0;$
|
||
$t=\dfrac{1}{x^2};$
|
||
$t\to+\infty;$
|
||
$\dfrac{e^{-1/x^2}}{x^{100}}=\dfrac{e^{-t}}{(1/t)^{50}}=\dfrac{t^{50}}{e^{t}};$
|
||
$\text{Неопределённость } \frac{\infty}{\infty}$
|
||
|
||
$\lim_{ t \to +\infty }\dfrac{t^{50}}{e^{t}}=\lim_{ t \to +\infty }\dfrac{(t^{50})'}{(e^{t})'}=\lim_{ t \to +\infty }\dfrac{50t^{49}}{e^{t}}=\dots=\lim_{ t \to +\infty }\dfrac{50!}{e^{t}}=0;$
|
||
$\lim_{ x \to 0 }\dfrac{e^{- \frac{1}{x^{2}}}}{x^{100}}=0;$
|
||
|
||
$\text{Ответ: }0$
|
||
|
||
## 112 номер – Д 1339
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ x \to +\infty } x^{2}e^{-0,01x}$
|
||
|
||
### Решение:
|
||
|
||
$x\to+\infty;$
|
||
$x^{2}e^{-0,01x}=\dfrac{x^2}{e^{0,01x}};$
|
||
$\dfrac{x^2}{e^{0,01x}}=\dfrac{\infty}{\infty};$
|
||
|
||
$\lim_{ x \to +\infty }\dfrac{x^2}{e^{0,01x}}=\lim_{ x \to +\infty }\dfrac{(x^2)'}{(e^{0,01x})'}=\lim_{ x \to +\infty }\dfrac{2x}{0,01e^{0,01x}}=\dfrac{\infty}{\infty};$
|
||
$\lim_{ x \to +\infty }\dfrac{2x}{0,01e^{0,01x}}=\lim_{ x \to +\infty }\dfrac{(2x)'}{(0,01e^{0,01x})'}=\lim_{ x \to +\infty }\dfrac{2}{0,0001e^{0,01x}}=0;$
|
||
|
||
$\text{Ответ: }0$
|
||
|
||
## 115 номер – Д 1342
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ x \to +0 }x^{x}$
|
||
|
||
### Решение:
|
||
|
||
$x\to0+;$
|
||
$x^{x}=0^{0};$
|
||
$\text{Неопределённость }0^{0}$
|
||
|
||
$y=x^x;$
|
||
$\ln y=x\ln x=\dfrac{\ln x}{1/x};$
|
||
$\dfrac{\ln x}{1/x}=\dfrac{-\infty}{\infty};$
|
||
$\lim_{ x \to 0+ }\ln y=\lim_{ x \to 0+ }\dfrac{(\ln x)'}{(1/x)'}=\lim_{ x \to 0+ }\dfrac{1/x}{-1/x^2}=\lim_{ x \to 0+ }(-x)=0;$
|
||
$\lim_{ x \to 0+ }y=e^{0}=1;$
|
||
|
||
$\text{Ответ: }1$
|
||
|
||
## 116 номер – Д 1343
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ x \to 0 }x^{x^{x}-1}$
|
||
|
||
### Решение:
|
||
|
||
$x\to0+;$
|
||
$x^{x^{x}-1}=0^{0};$
|
||
$\text{Неопределённость }0^{0}$
|
||
|
||
$y=x^{x^{x}-1};$
|
||
$\ln y=(x^x-1)\ln x=\dfrac{x^x-1}{1/\ln x};$
|
||
$\dfrac{x^x-1}{1/\ln x}=\dfrac{0}{0};$
|
||
|
||
$(x^x)'=x^x(\ln x+1);$
|
||
$\lim_{ x \to 0+ }\ln y=\lim_{ x \to 0+ }\dfrac{(x^x-1)'}{(1/\ln x)'}=\lim_{ x \to 0+ }\dfrac{x^x(\ln x+1)}{-1/(x(\ln x)^2)}=\lim_{ x \to 0+ }(-x^{x+1}(\ln x+1)(\ln x)^2);$
|
||
|
||
$\lim_{ x \to 0+ }x^x=1;$
|
||
$\lim_{ x \to 0+ }x^{x+1}=\lim_{ x \to 0+ }x\cdot x^x=0;$
|
||
|
||
$\lim_{ x \to 0+ }x(\ln x)^2=\lim_{ x \to 0+ }\dfrac{(\ln x)^2}{1/x}=\dfrac{\infty}{\infty};$
|
||
$\lim_{ x \to 0+ }\dfrac{(\ln x)^2}{1/x}=\lim_{ x \to 0+ }\dfrac{(2\ln x\cdot\frac{1}{x})}{(-1/x^2)}=\lim_{ x \to 0+ }(-2x\ln x);$
|
||
$\lim_{ x \to 0+ }x\ln x=\lim_{ x \to 0+ }\dfrac{\ln x}{1/x}=\dfrac{-\infty}{\infty};$
|
||
$\lim_{ x \to 0+ }\dfrac{\ln x}{1/x}=\lim_{ x \to 0+ }\dfrac{1/x}{-1/x^2}=\lim_{ x \to 0+ }(-x)=0;$
|
||
$\lim_{ x \to 0+ }x(\ln x)^2=0;$
|
||
|
||
$\lim_{ x \to 0+ }x(\ln x)^3=\lim_{ x \to 0+ }\dfrac{(\ln x)^3}{1/x}=\dfrac{-\infty}{\infty};$
|
||
$\lim_{ x \to 0+ }\dfrac{(\ln x)^3}{1/x}=\lim_{ x \to 0+ }\dfrac{(3(\ln x)^2\cdot\frac{1}{x})}{(-1/x^2)}=\lim_{ x \to 0+ }(-3x(\ln x)^2)=0;$
|
||
|
||
$\lim_{ x \to 0+ }x(\ln x+1)(\ln x)^2=\lim_{ x \to 0+ }(x(\ln x)^3+x(\ln x)^2)=0;$
|
||
$\lim_{ x \to 0+ }\ln y=\lim_{ x \to 0+ }(-x^x)\cdot\lim_{ x \to 0+ }x(\ln x+1)(\ln x)^2= -1\cdot0=0;$
|
||
$\lim_{ x \to 0+ }y=e^{0}=1;$
|
||
|
||
$\text{Ответ: }1$
|
||
|
||
## 119 номер – Д 1348
|
||
|
||
### Пример:
|
||
|
||
$\lim_{ x \to \frac{\pi}{4} }(\tan x)^{\tan 2x}$
|
||
|
||
### Решение:
|
||
|
||
$x\to\dfrac{\pi}{4};$
|
||
$\tan x\to1;$
|
||
$\tan2x\to\pm\infty;$
|
||
$\text{Неопределённость }1^{\infty}$
|
||
|
||
$y=(\tan x)^{\tan2x};$
|
||
$\ln y=\tan2x\cdot\ln(\tan x)=\dfrac{\ln(\tan x)}{\cot2x};$
|
||
$\dfrac{\ln(\tan x)}{\cot2x}=\dfrac{0}{0};$
|
||
|
||
$\lim_{ x \to \frac{\pi}{4} }\ln y=\lim_{ x \to \frac{\pi}{4} }\dfrac{(\ln(\tan x))'}{(\cot2x)'}=\lim_{ x \to \frac{\pi}{4} }\dfrac{\dfrac{\sec^2x}{\tan x}}{-2\csc^2(2x)}=\lim_{ x \to \frac{\pi}{4} }(-\dfrac{\sin^2(2x)}{2\sin x\cos x});$
|
||
$\sin^2(2x)=4\sin^2x\cos^2x;$
|
||
$-\dfrac{\sin^2(2x)}{2\sin x\cos x}=-2\sin x\cos x=-\sin2x;$
|
||
$\lim_{ x \to \frac{\pi}{4} }\ln y=-\sin\dfrac{\pi}{2}=-1;$
|
||
$\lim_{ x \to \frac{\pi}{4} }y=e^{-1};$
|
||
|
||
$\text{Ответ: }e^{-1}$
|