Question

Помогите с сочем.Умоляю.Дам 20 баллов за правильное решение

Answer 1

Ответ:

$1)\ \ \lim\limits_{x \to \infty}\dfrac{5x^2-3x+7}{9x^2+2x-5}=\lim\limits_{x \to \infty}\dfrac{5x^2}{9x^2}=\dfrac{5}{9}\\\\(\ (a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_0)\sim a_{n}x^{n}\ ,\ x\to \infty \ )\\\\\\\lim\limits_{x \to 0}\dfrac{sin7x}{3x}=\lim\limits_{x \to 0}\dfrac{7x}{3x}=\dfrac{7}{3}\ \ \ \ \ \ (sina\sim a\ ,\ a\to 0\ )$

$2a)\ \ y=arcsin3x\ \ \to \ \ 3x=siny\ \ ,\ \ x=\varphi (y)=\dfrac{1}{3}\, siny\\\\f'(x)=\dfrac{1}{\varphi'(y)}\ \ \to \ \ \varphi'(y)=\dfrac{1}{f'(x)}=\dfrac{1}{\frac{3}{\sqrt{1-9x^2}}}=\dfrac{\sqrt{1-9x^2}}{3}=\dfrac{\sqrt{1-9\cdot \frac{1}{9}sin^2y}}{3}=\\\\=\dfrac{\sqrt{1-sin^2y}}{3\frac{x}{y} }=\dfrac{cosy}{3}=\dfrac{1}{3}\, cosy$

$P.S.\ \ \ (arcsin3x)'=f'(x)=\dfrac{1}{\varphi '(y)}=\dfrac{1}{\frac{1}{3}\, cosy}=\dfrac{3}{\sqrt{1-sin^2y}}=\\\\\\=\dfrac{3}{\sqrt{1-sin^2(arcsin3x)}}=\dfrac{3}{\sqrt{1-(3x)^2}}=\dfrac{3}{\sqrt{1-9x^2}}$

$b)\ \ f(x)=x^3-arccos2x\\\\f'(x)=3x^2+\dfrac{2}{\sqrt{1-4x^2}}\\\\\varphi '(y)=\dfrac{1}{f'(x)}=\dfrac{1}{3x^2+\frac{2}{\sqrt{1-4x^2}}}=\dfrac{\sqrt{1-4x^2}}{3x^2\sqrt{1-4x^2}+2}$