Question

Найти асимптоты
1)f(x)=x^3/x^2-9
2)f(x)=x/x^2-x-2

Answer 1

$1)\; \; f(x)=\frac{x^3}{x^2-9}\; \; ,\; \; \; f(x)=\frac{x^3}{(x-3)(x+3)}\\\\a)\; \; \lim\limits _{x \to 3}\frac{x^3}{(x-3)(x+3)}=[\frac{27}{0}]=\+\infty \; \; \; \to \; \; \; \; x=3\; \; asimptota\\\\b)\; \; \lim\limits _{x \to -3}\frac{x^3}{(x-3)(x+3}=[\frac{-27}{0}]=-\infty \; \; \; \to \; \; \; x=-3\; \; asimptota\\\\c)\; \; y=kx+b\\\\k=\lim\limits _{x \to \infty}\frac{f(x)}{x}=\lim\limits _{x \to \infty}\frac{x^3}{x^3-9}=1$

$b=\lim\limits _{x \to \infty}(y-kx)=\lim\limits _{x \to \infty}(\frac{x^3}{x^2-9}-x)=\lim\limits _{x \to \infty}\frac{x^3-x^3+9x}{x^2-8}=\lim\limits _{x \to \infty}\frac{9x}{x^2-9}=0\\\\y=x\; \; asimptota\\\\\\2)\; \; f(x)=\frac{x}{x^2-x-2}\; \; ,\; \; \; f(x)=\frac{x}{(x-2)(x+1)}\\\\\lim\limits _{x \to 2}\frac{x}{(x-2)(x+1)}=[\frac{2}{0}]=+\infty \; \; \; \to \; \; \; x=2\; \; asimptota$

$\lim\limits _{x \to -1}\frac{x}{(x-2)(x+1)}=[\frac{-1}{0}]=-\infty \; \; \; \to \; \; \; x=-1\; \; asimptota\\\\y=kx+b:\; k=\lim\limits _{x \to \infty}\frac{x}{x\cdot (x^2-x-2)}=\lim\limits _{x \to \infty}\frac{1}{x^2-x-2}=0\; ,\; \; k=0\\\\b=\lim\limits _{x \to \infty}\frac{x}{x^2-x-2}=0\\\\y=0\; \; \; gorizontalnaya\; \; asimptota$