Асимптоты и графики функций

$y = \frac{1}{1-x^{2}}$;

$y = \frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+2}$;

$y = \frac{x^{3}}{6 x^{2}-8-x^{4}}$;

$y = \frac{x}{1+x^{2}}$;

$y = \frac{x^{2}+2 x+1}{x^{2}+1}$

$y = \sqrt{\frac{x-2}{x+2}}$

$y = \sqrt{x+1}-\sqrt{x-1}$;

$y = (2-x)^{2 / 3}-(2+x)^{2 / 3}$.

$y = x+\frac{1}{x^{2}}$

$y = x+\frac{x^{2}}{x^{2}+1}$

$y = \frac{x^{2}+8 x-6}{x}$

$y = \frac{x^{2}}{x+4}$;

$y = \frac{x^{2}}{\left|x\right|+1}$;

$y = \frac{2 x^{4}+x^{3}+1}{x^{3}}$;

$y = \frac{x^{3}}{(x+1)^{2}}$;

$y = \frac{x^{2}-2 x+3}{x+2}$;

$y = \frac{x^{3}-3 a x^{2}+a^{3}}{x^{2}-3 b x+2 b^{2}}$;

$y = \frac{x^{5}}{x^{4}+1}$.

$y = \sqrt{x^{2}-4}$;

$y = \sqrt{x^{2}+3 x-1}$

$y = \sqrt[3]{x^{3}-6 x}$

$y = \sqrt[3]{x^{3}+x^{2}}$

$y = \sqrt{\frac{x^{3}}{x-2}}$

$y = x \sqrt{\frac{x}{x+4}}$

$y = \sqrt{x^{4}+x^{3}}-\sqrt{x^{4}-x^{3}}$;

$y = \sqrt{x^{2}-1}-x$;

$y = x+\sqrt{4 x^{2}+1}$.

$y = e^{-1 / x}$;

$y = 0,1^{x^{2}}$;

$y = 2^{-1 / x^{2}}$;

$y = 2^{x} / x$;

$y = x^{2} e^{x}$;

$y = e^{1 / x}-x$;

$y = 1-x e^{-1 /\left|x\right|-1 / x}$;

$y = 1+x e^{2 / x}$;

$y = x 2^{1 / x^{2}}$

$y = \left|e^{x}-1\right|$;

$y = \left|x+2\right| e^{-1 / x}$.

$y = (1+1 / x)^{x}$;

$y = x(1-1 / x)^{x}$;

$y = \operatorname {th} x$;

$y = \operatorname {cth} x$;

$y = \operatorname {th}^{2} x$;

$y = x \operatorname {th} x$;

$y = 2 x+\operatorname {cth} x$;

$y = x \frac{1-e^{-x}}{1+e^{-x}}$.

$y = \log_{3}\left(4-x^{2}\right)$;

$y = \log_{0,5}\left(2 x^{2}-3 x+1\right)$;

$y = \lg \sin 2 x$;

$y = x+\frac{\ln x}{x}$

$y = \ln \left(1+e^{x}\right)$

$y = x \lg \left(e+\frac{1}{x}\right)$.

$y = \frac{1}{\cos (x-\pi / 6)}$;

$y = \operatorname {cosec} 2 x$;

$y = \frac{\cos x}{x}$;

$y = x+\frac{\sin x}{2 x}$;

$y = x \sin \frac{1}{x}$;

$y = 2+\cos \frac{2}{x}$;

$y = \operatorname {arcctg} x$;

$y = \operatorname {arctg} \frac{1}{x}$

$y = \arcsin \frac{1}{x^{2}}$

$y = \frac{x}{2}+\arccos \frac{1}{x+1}$;

$y = 2 x-\operatorname {arctg} \frac{x}{2}$;

$y = x \operatorname {arcctg} x$;

$y = x \operatorname {arctg} x$;

$y = \frac{1}{\operatorname {arcctg}(1 / x)}$.

$f(x) = \frac{x(x+2)}{x+1}, x > -1$;

$f(x) = \frac{2 x^{3}}{x^{2}+1}$

$f(x) = 1-2 \sqrt[3]{x^{3}+3 x}$

$f(x) = \operatorname {th} x$;

$f(x) = \operatorname {cth} x$

$f(x) = 2 x+\operatorname {arctg} x$.

Функция $f$ определена на интервале $(a ;+\infty )$. Доказать, что для того, чтобы прямая $y = k x+b$ была асимптотой графика функции $f$ при $x \rightarrow +\infty$, необходимо и достаточно, чтобы расстояние $\rho (x)$ от точки $(x ; f(x))$ до этой прямой стремилось к нулю при $x \rightarrow +\infty$.

Доказать, что в случае вертикальной асимптоты необходимость предыдущего утверждения верна, а достаточность неверна.

$y = \frac{x^{3 / 2}}{\sqrt{1-x}}$;

$y = x \sqrt{4-x^{2}}$;

$y = \sqrt{x^{2}-x^{4}}$;

$y = \sqrt[4]{9 x^{2}-x^{4}}$;

$y = \sqrt[3]{\frac{x^{2}}{1+x^{2}}}$;

$y = \sqrt[4]{\frac{x^{4}}{3+x^{4}}}$;

$y = \frac{x}{\sqrt{x^{2}-1}}$;

$y = \sqrt{1-\frac{1}{x}}$

$y = x^{3 / 2}-4 x^{1 / 2}$.

$\sqrt{f(x)} = \sqrt{a\left(x-x_{0}\right)}+o\left(\sqrt{\left|x-x_{0}\right|}\right)$ при $a\left(x-x_{0}\right) \geqslant 0$, т. е. графики функций $\sqrt{f(x)}$ и $\sqrt{a\left(x-x_{0}\right)}$ "сближаются" при $x \rightarrow x_{0}$;

если $\alpha > 0$, то

при $a\left(x-x_{0}\right) > 0$, т. е. графики функций $(f(x))^{\alpha }$ и $\left(a\left(x-x_{0}\right)\right)^{\alpha }$ "сближаются" при $x \rightarrow x_{0}$.

Функция $f$ определена в окрестности точки $x_{0}$ и

Доказать, что

т. е. графики функций $\frac{1}{f(x)}$ и $\frac{1}{a\left(x-x_{0}\right)}$ "сближаются" при $x \rightarrow x_{0}$.

Проверить, что функция $f(x) = x+x^{4 / 3}$ удовлетворяет условию

но равенство неверно.

$f_{1}(x) = \sqrt{x^{3}+a x^{2}+b x+c}, f_{2}(x) = x^{3 / 2}, x \rightarrow +\infty$;

$f_{1}(x) = \operatorname {ch} x, f_{2}(x) = e^{x} / 2, x \rightarrow +\infty$;

$f_{1}(x) = \operatorname {sh} x, f_{2}(x) = -e^{-x} / 2, x \rightarrow -\infty$;

$f_{1}(x) = \operatorname {ctg} x, f_{2}(x) = 1 / x, x \rightarrow 0$

$f_{1}(x) = \frac{x^{2}}{\sqrt{x^{2}-1}}, f_{2}(x) = \frac{1}{\sqrt{2(x-1)}}, x \rightarrow 1+0$.

$f(x) = x+\sqrt{x}, g(x) = x+\sqrt{\frac{x}{x-1}}$

$f(x) = \ln \left(e^{x} \cdot x^{1 / \sqrt{x}}\right), g(x) = \ln \frac{e^{x}}{x+2}$

$f(x) = \frac{(x+1) \ln x+1}{\ln x+1}, g(x) = \frac{(x+1) \ln x+x}{\ln x+1}$;

$f(x) = \sqrt[3]{x^{3}+x^{2}}, g(x) = \sqrt[4]{x^{4}+b x}$.

$y = \frac{x^{2}+1}{x}$

$y = \frac{x^{3}+1}{x^{2}}$;

$y = \frac{(x+1)^{3}}{(x+2)^{2}}$

$y = \frac{x^{2}}{\left|x+1\right|}$

$y = \frac{\sqrt{x^{4}+1}}{x}$;

$y = \frac{x^{3}+2 \sin x}{x^{2}+1}$.

$y = \frac{2-4 x^{2}}{1-4 x^{2}}$; 2

$y = \frac{x-1}{x^{2}-2 x}$;

$y = \frac{x^{3}}{1-x^{3}}$;

$y = \frac{(x-3)^{2}}{4 x-4}$

$y = \frac{x^{3}}{1+x^{2}}$

$y = \sqrt[5]{\frac{x}{x-2}}$

$y = \sqrt[3]{1-x^{3}}$;

$y = \frac{x^{2}}{\sqrt{\left|x^{2}-1\right|}}$

$y = 1-\sqrt{4 x^{2}+1}$;

$y = \sqrt{x^{2}+4 x+5}-2$

$y = \sqrt[3]{x^{3}-x^{2}}$

$y = e^{1 /(3+x)}$;

$y = \ln \left(4-x^{2}\right)$;

$y = \frac{e^{x}}{e^{x}-1}$

$y = \log_{x^{2}-6 x+7} 2$;

$y = \ln \left(1+2 e^{x}\right)$;

$y = \ln \cos x$;

$y = \ln \operatorname {arctg} x$;

$y = e^{1 / \sin x}$.

$y = x+\frac{\sin x}{x}$;

$y = x^{2} \sin \frac{1}{x}$

$y = (x+1)^{2} \sin \frac{2}{x}$;

$y = x+\operatorname {arctg} x$

$y = (x+1) \operatorname {arcctg} x$;

$y = x \arccos \frac{1}{x}$

$y = 3 x+\operatorname {arctg} 5 x$;

$y = \operatorname {arctg} \frac{1}{x}-x$;

$y = x^{2} \operatorname {arctg} \frac{1}{x}$.

$y = \frac{x-\mathbb {E}\left[x\right]}{\mathbb {E}\left[x\right]}$;

$y = \frac{\sin x}{x^{2}-x}$

$y = \frac{\mathbb {E}\left[x^{3}\right]}{x^{2}+1}$

$y = \frac{\sin x}{E(x / \pi )+0,5}$.

$x = \frac{3 a t}{1+t^{3}}, y = \frac{3 a t^{2}}{1+t^{2}}, a > 0$;

$x = \frac{t-8}{t^{2}-4}, y = \frac{3}{t\left(t^{2}-4\right)}$

$x = \frac{a t^{4}}{1-t^{3}}, y = \frac{a t^{3}}{1-t^{3}}, a > 0$;

$x = \frac{t^{3}}{t^{2}+1}, y = \frac{t^{3}-2 t^{2}}{t^{2}+1}$

$x = \frac{5 a t^{2}}{1+t^{5}}, y = \frac{5 a t^{3}}{1+t^{5}}, a > 0$

$x = t^{3}+3 t+1, y = t^{3}-3 t+1$

$x = t^{3}-3 \pi , y = t^{3}-\operatorname {arctg} t$;

$x = t+\sin t, y = t+\cos t$

$x = t e^{t}, y = t e^{-t}$

$x = t \ln t, y = t \ln (t+1)$

$x = \frac{2 e^{t}}{t-1}, y = \frac{t e^{t}}{t-1}$;

$x = t+e^{-t}, y = 2 t+e^{-t}$.

$x = \frac{1}{t}, y = \frac{t}{t+1}$;

$x = \frac{2 t}{1-t^{2}}, y = \frac{t^{2}}{1-t^{2}}$;

$x = a \operatorname {ch} t, y = b \operatorname {sh} t$;

$x = 2 \cos t, y = \operatorname {tg} 2 t$;

$x = \frac{1}{\sin t}, y = \frac{1}{\sin 2 t}$;

$x = \frac{t}{t^{2}+1}, y = \frac{t^{2}}{t-1}$

$x = \frac{2+t^{2}}{1+t^{2}}, y = \frac{t^{3}}{1+t^{2}}$

$x = \frac{t^{2}}{t-1}, y = \frac{t}{t^{2}-1}$.

$r = \sqrt[4]{\frac{24}{\sin 2 \varphi (1-\sin 2 \varphi )}}$

$r = \sqrt[4]{-\frac{4}{\sin 4 \varphi }}$

$r = \frac{2 a \sin^{2} \varphi }{\cos \varphi }$

$r = \frac{4 \cos^{2} \varphi \sin \varphi }{\cos 2 \varphi }$ (скифоида);

$r = \frac{3 \cos^{2} \varphi }{\cos^{3} \varphi +\sin^{3} \varphi }$

$r = \frac{\sin^{2} \varphi }{\cos^{2} \varphi (\cos \varphi -2 \sin \varphi )}$.

$r = \frac{\pi }{\varphi -\pi / 4}$;

$r = \sqrt{\frac{\pi }{\varphi }}$ (жезл);

$\varphi = \frac{r}{r-1}, r > 1$;

$\varphi = \frac{r}{r^{2}+1}, r \geqslant 0$

$r = 2 \operatorname {tg} \varphi$

$r = \frac{p}{1-\varepsilon \cos \varphi }, p > 0, \varepsilon > 1$;

$r = -\frac{p}{1+\varepsilon \cos \varphi }, p > 0, \varepsilon > 1$;

$r = 2\left|1-\operatorname {tg} \varphi \right|$;

$r = \frac{2}{\left|\sin 2 \varphi \right|}$;

$r = \frac{a}{\cos 3 \varphi }, a > 0$;

$r = \frac{a}{\sqrt{\cos 3 \varphi }}, a > 0$;

$r = a \frac{\operatorname {th} \varphi }{\varphi -1}, a > 0$;

$r = 2 R \operatorname {tg} \varphi \sin \varphi , R > 0$ (циссоида);

$r = \frac{a}{\cos \varphi } \pm a \operatorname {tg} \varphi , a > 0$ (строфоида);

$r = a \frac{\cos 2 \varphi }{\cos \varphi }$;

$r = \sqrt[3]{\frac{2 \cos \varphi }{\cos^{2} 2 \varphi }}$;

$r = \frac{\sqrt{2 \sin 2 \varphi }}{\left|\cos 2 \varphi \right|}$;

$r = \sqrt{-\frac{\operatorname {tg} 2 \varphi }{2}}$.

$x^{2} y^{2}+x-2 y = 0$;

$x^{2} y^{2}+y = 1$;

$(x-1)(x-2) y^{2} = x^{2}$;

$x^{2} y+x y^{2} = 1$;

$y^{3}-x^{3} = 6 x^{2}$

$y^{2}(a-x) = x^{2}(a+x), a > 0$;

$x^{3}-3 x y^{2} = 2$;

$\left(x^{2}-1\right)^{3}-x^{4} y^{2} = 0$;

$x^{4}-y^{4} = x^{2}-2 y^{2}$;

$x y(x-y)+x+y = 0$;

$x y(x+y)+x^{2} = 2 y^{2}$;

$\left(x^{2}-1\right) y^{2} = x^{2}\left(x^{2}-4\right)$;

$y^{3}-x^{3}+y-2 x = 0$;

$x^{4}-y^{4} = 4 x^{2} y$;

$x^{4}-2 x^{2} y^{2}+y^{3} = 0$;

$\left(x+\frac{a}{3}\right)\left(y^{2}-\frac{x^{2}}{3}\right)+\frac{4}{9} a x^{2} = 0, a > 0$;

$x^{y} = y^{x}$.

$x^{2} / 9-y^{2} / 16 = 1$;

$(y+x+1)^{2} = x^{2}+1$;

$2 x^{2}-5 x y+2 y^{2}+x+y-3 = 0$;

$4 x^{2}+9 y^{2} = x^{2} y^{2}$

$y^{3}+x^{3} = 8$;

$y^{2}\left(x^{2}+1\right) = x^{2}\left(x^{2}-1\right)$;

$y^{2}(4-x) = x^{3}$ (циссоида);

$4 x^{4 / 3}-y^{4 / 3} = 1$;

$x^{2} y^{2}+y^{4} = 4 x^{2}$;

$x y\left(x^{2}-y^{2}\right)+1 = 0$;

$\left(x^{2}-y^{2}\right)^{2}+4 x y = 0$

$\left(x^{2}+y^{2}\right)(y-1)^{2}-y^{2} = 0$ (конхоида).