Функције – асимптоте функција 1

Задаци

Текстови задатака објашњених у видео лекцији.

Одредити вертикалну асимптоту графика функције:

пр.1)   $f(x) = lnx$

пр.2)   $f(x) = ctgx$

пр.3)   $f(x) = \frac{1}{{{x^2} - 9}}$

пр.1)   $f(x) = lnx$

$f\left( x \right){\text{ }} = {\text{ }}lnx$

$Df:x \in \left( {0; + \infty } \right)$

$B.A.:\mathop {lim}\limits_{x \to {0^ + }} lnx = - \infty$

$B.A.:x = 0$

пр.2)   $f(x) = ctgx$

$f\left( x \right){\text{ }} = {\text{ }}ctgx$

$Df:\mathbb{R}\backslash \left\{ {k\pi ,k \in \mathbb{Z}} \right\}$

$\mathop {\lim }\limits_{x \to {\pi ^ - }} ctgx = - \infty$

$\mathop {\lim }\limits_{x \to {\pi ^ + }} ctgx = + \infty$

$B.A.:\boxed{x = \pi }$

пр.3) 

$$\begin{aligned}&f(x) = \frac{1}{{{x^2} - 9}} \\ &Df:{x^2} - 9 \ne 0 \\ &\left( {x - 3} \right)\left( {x + 3} \right) \ne 0 \\ &x \ne 3;x \ne - 3 \\ &Df:\mathbb{R}\backslash \left\{ { - 3;3} \right\}\end{aligned}$$

$$\begin{aligned}&\mathop {\lim }\limits_{x \to - {3^ - }} \frac{1}{{{x^2} - 9}} = \frac{1}{{{9^ + } - 9}} = \frac{1}{{{0^ + }}} = \infty \\ &\mathop {\lim }\limits_{x \to - {3^ + }} \frac{1}{{{x^2} - 9}} = \frac{1}{{{9^ - } - 9}} = \frac{1}{{{0^ - }}} = - \infty \\ &\mathop {\lim }\limits_{x \to {3^ - }} \frac{1}{{{x^2} - 9}} = \frac{1}{{{9^ - } - 9}} = \frac{1}{{{0^ - }}} = - \infty \\ &\mathop {\lim }\limits_{x \to {3^ + }} \frac{1}{{{x^2} - 9}} = \frac{1}{{{9^ + } - 9}} = \frac{1}{{{0^ + }}} = \infty\end{aligned}$$

$$B.A.:\boxed{x = 3;x = - 3}$$