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)
\[\begin{gathered}
f(x) = \frac{1}{{{x^2} - 9}} \hfill \\
Df:{x^2} - 9 \ne 0 \hfill \\
\left( {x - 3} \right)\left( {x + 3} \right) \ne 0 \hfill \\
x \ne 3;x \ne - 3 \hfill \\
Df:\mathbb{R}\backslash \left\{ { - 3;3} \right\} \hfill \\
\end{gathered} \]
\[\begin{gathered}
\mathop {\lim }\limits_{x \to - {3^ - }} \frac{1}{{{x^2} - 9}} = \frac{1}{{{9^ + } - 9}} = \frac{1}{{{0^ + }}} = \infty \hfill \\
\mathop {\lim }\limits_{x \to - {3^ + }} \frac{1}{{{x^2} - 9}} = \frac{1}{{{9^ - } - 9}} = \frac{1}{{{0^ - }}} = - \infty \hfill \\
\mathop {\lim }\limits_{x \to {3^ - }} \frac{1}{{{x^2} - 9}} = \frac{1}{{{9^ - } - 9}} = \frac{1}{{{0^ - }}} = - \infty \hfill \\
\mathop {\lim }\limits_{x \to {3^ + }} \frac{1}{{{x^2} - 9}} = \frac{1}{{{9^ + } - 9}} = \frac{1}{{{0^ + }}} = \infty \hfill \\
\end{gathered} \]
\[B.A.:\boxed{x = 3;x = - 3}\]
