Question

100 балов

задание во вложение

Answer 1

Ответ:

а) $\boldsymbol{\boxed{\lim_{x \to 0} \dfrac{\sin^{2} x}{1 - \cos 4x} = \frac{1}{8} }}$

б) $\boldsymbol{\boxed{ \lim_{x \to -4} (x + 5)^{\dfrac{3}{x + 4} } = e^{3}}}$

в) $\boldsymbol{\boxed{ \lim_{x \to 0} \dfrac{1 -\cos^{3} x}{x \sin 3x} = \frac{1}{2} }}$

г) $\boldsymbol{\boxed{\lim_{x \to \infty} \bigg(\frac{x + 6}{x + 2} \bigg)^{3x - 6} = e^{12} }}$

Примечание:

По теореме:

$\boxed{ \lim_{x \to a} u(x)^{v(x)} = \bigg[1^{\infty} \bigg ] = \bigg e^{\displaystyle \lim_{x \to a} (v(x)(u(x) - 1)) } }$

Второй замечательный предел:

$\boxed{ \lim_{x \to \infty} \bigg(1 + \frac{1}{x} \bigg)^{x} = e}$

По таблице производных:

$\boxed{(x^{n})' = nx^{n - 1}}$

$\boxed{(\sin x)' = \cos x}$

$\boxed{(\cos x)' = -\sin x }$

$\boxed{C' = 0}$, где $C \in \mathbb R$

Правила дифференцирования:

$(f \pm g)' = f' \pm g'$

$(fg)' = f'g + fg'$

$\bigg(\dfrac{f}{g} \bigg)' = \dfrac{f'g - fg'}{g^{2}}$

$f(g) = g'f'(g)$

$(kf)' = k(f')$, где $k \in \mathbb R$

$f,g \ -$ функции одной переменной

Пошаговое объяснение:

а)

$\displaystyle \lim_{x \to 0} \dfrac{\sin^{2} x}{1 - \cos 4x} = \bigg [ \frac{0}{0} \bigg ] = \lim_{x \to 0} \dfrac{(\sin^{2} x)'}{(1 - \cos 4x)'} = \lim_{x \to 0} \dfrac{2 \sin x \cos x}{4 \sin 4x} =$

$\displaystyle = \lim_{x \to 0} \dfrac{\sin 2x}{4 \cdot 2\sin 2x \cos 2x} =\lim_{x \to 0} \dfrac{1}{8 \cos 2x} = \dfrac{1}{8 \cdot \cos (2 \cdot 0)} = \dfrac{1}{8 \cdot \cos 0}= \dfrac{1}{8 \cdot 1} = \frac{1}{8}$

б)

$\displaystyle \lim_{x \to -4} (x + 5)^{\dfrac{3}{x + 4} } = \bigg[1^{\infty} \bigg ] = \Bigg e^{\displaystyle \lim_{x \to -4} \bigg( \dfrac{3}{x + 4}(x + 5 - 1) \bigg ) } = \Bigg e^{\displaystyle \lim_{x \to -4} \bigg( \dfrac{3(x + 4)}{(x + 4)} \bigg ) } =$

$=\bigg e^{\displaystyle \lim_{x \to -4} 3 } = e^{3}$

в)

$\displaystyle \lim_{x \to 0} \dfrac{1 -\cos^{3} x}{x \sin 3x} = \bigg [ \frac{0}{0} \bigg ] = \lim_{x \to 0} \dfrac{(1 -\cos^{3} x)'}{(x \sin 3x)'} = \lim_{x \to 0} \dfrac{3\sin x\cos^{2} x}{(x)'\sin 3x + x(\sin 3x)'} =$

$\displaystyle = \lim_{x \to 0} \dfrac{3\sin x\cos^{2} x}{\sin 3x + 3x \cos 3x} = \bigg [ \frac{0}{0} \bigg ] = \lim_{x \to 0} \dfrac{(3\sin x\cos^{2} x)'}{(\sin 3x + 3x \cos 3x)'}=$

$\displaystyle = 3 \lim_{x \to 0} \dfrac{(\sin x)'\cos^{2} x + \sin x(\cos^{2} x)' }{(\sin 3x)' + (3x \cos 3x)'}= 3 \lim_{x \to 0} \dfrac{\cos^{3} x - 2\sin^{2} x\cos x }{(3x)'\cos 3x + ((3x)' \cos 3x + 3x(\cos 3x)')}=$

$\displaystyle =3 \lim_{x \to 0} \dfrac{\cos^{3} x - 2\sin^{2} x\cos x }{3\cos 3x + (3 \cos 3x - (3x)'3x\sin 3x)}= 3 \lim_{x \to 0} \dfrac{\cos^{3} x - 2\sin^{2} x\cos x }{3\cos 3x+ (3 \cos 3x - 9x\sin 3x)}=$

$\displaystyle =3 \lim_{x \to 0} \dfrac{\cos^{3} x - 2\sin^{2} x\cos x }{3\cos 3x + 3 \cos 3x - 9x\sin 3x}= 3 \lim_{x \to 0} \dfrac{\cos^{3} x - 2\sin^{2} x\cos x }{6\cos 3x- 9x\sin 3x}=$

$\displaystyle= \frac{3}{3} \lim_{x \to 0} \dfrac{\cos^{3} x - 2\sin^{2} x\cos x }{2\cos 3x- 3x\sin 3x}= \lim_{x \to 0} \dfrac{\cos^{3} x - 2\sin^{2} x\cos x }{2\cos 3x- 3x\sin 3x}=$

$= \dfrac{\cos^{3} 0 - 2\sin^{2} 0\cos 0 }{2\cos (3\cdot 0) - 3 \cdot 0\sin (3 \cdot 0)}=\dfrac{1 - 0}{2 \cdot 1 - 0} = \dfrac{1}{2}$

г)

$\displaystyle \lim_{x \to \infty} \bigg(\frac{x + 6}{x + 2} \bigg)^{3x - 6} = \lim_{x \to \infty} \frac{\bigg(\dfrac{x + 6}{x + 2} \bigg)^{3x} }{\bigg(\dfrac{x + 6}{x + 2} \bigg)^{6}} = \lim_{x \to \infty} \frac{ \Bigg( \bigg(\dfrac{x + 4 + 2}{x + 2} \bigg)^{x} \Bigg)^{3} }{\bigg(\dfrac{x + 2 + 4}{x + 2} \bigg)^{6}} =$

$\displaystyle = \frac{ \lim_{x \to \infty} \Bigg( \dfrac{x + 2 }{x + 2} + \dfrac{ 4}{x + 2} \bigg)^{x} \Bigg)^{3} }{ \lim_{x \to \infty}\bigg(\dfrac{x + 2 }{x + 2} + \dfrac{ 4}{x + 2} \bigg)^{6}} = \frac{ \lim_{x \to \infty} \Bigg( 1 + \dfrac{1}{0,25(x + 2)} \bigg)^{x} \Bigg)^{3} }{ \lim_{x \to \infty}\bigg(1 + \dfrac{4}{x + 2} \bigg)^{6}} =$

$\displaystyle = \frac{ \lim_{x \to \infty} \Bigg( 1 + \dfrac{1}{0,25x + 0,5} \bigg)^{x} \Bigg)^{3} }{1} = \lim_{x \to \infty} \Bigg( \Bigg( 1 + \dfrac{1}{0,25x + 0,5} \bigg)^{0,25x + 0,5 - 0,5} \Bigg)^{4}\Bigg)^{3}=$

$\displaystyle = \Bigg( \lim_{x \to \infty} \Bigg( 1 + \dfrac{1}{0,25x + 0,5} \bigg)^{0,25x + 0,5 - 0,5} \Bigg)^{12}=$

$\displaystyle = \frac{ \Bigg(\displaystyle\lim_{x \to \infty} \Bigg( 1 + \dfrac{1}{0,25x + 0,5} \bigg)^{0,25x + 0,5 }\Bigg)^{12}}{\Bigg(\displaystyle \lim_{x \to \infty} \Bigg( 1 + \dfrac{1}{0,25x + 0,5} \bigg)^{0,5 }\Bigg)^{12}} = \dfrac{e^{12}}{1} = e^{12}$