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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)   $f(x) = lnx$

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

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

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

$B.A.:x = 0$

пр.2)   $f(x) = ctgx$

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

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

439 png

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

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

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

пр.3) 

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

440 png

\[\begin{gathered}
\mathop {\lim }\limits_{x \to - {3^ - }} \frac{1}{{{x^2} - 9}} = \frac{1}{{{9^ + } - 9}} = \frac{1}{{{0^ + }}} = \infty \hfill \\
\mathop {\lim }\limits_{x \to - {3^ + }} \frac{1}{{{x^2} - 9}} = \frac{1}{{{9^ - } - 9}} = \frac{1}{{{0^ - }}} = - \infty \hfill \\
\mathop {\lim }\limits_{x \to {3^ - }} \frac{1}{{{x^2} - 9}} = \frac{1}{{{9^ - } - 9}} = \frac{1}{{{0^ - }}} = - \infty \hfill \\
\mathop {\lim }\limits_{x \to {3^ + }} \frac{1}{{{x^2} - 9}} = \frac{1}{{{9^ + } - 9}} = \frac{1}{{{0^ + }}} = \infty \hfill \\
\end{gathered} \]

\[B.A.:\boxed{x = 3;x =  - 3}\]

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