Функције – асимптоте функција 2

Текстови задатака објашњених у видео лекцији.

Одредити асимптоту графика функције:

пр.4)   $f(x) = {2^x}$

пр.5)   $f(x) = arctgx$

пр.6)   $f(x) = \frac{{2 - 2{x^2}}}{{{x^2} + 3}}$

пр.7)   $f(x) = \sqrt {{x^2} + 2}  - \sqrt {{x^2} - 5x}$

пр.4)   $f(x) = {2^x}$

$\mathop {\lim }\limits_{x \to  - \infty } {2^x} = 0$

$\mathop {\lim }\limits_{x \to  + \infty } {2^x} =  + \infty $

Хоризонтална асимптота:

$\boxed{\mathop {y = 0}\limits_{x \to  - \infty } }$

пр.5)   $f(x) = arctgx$

$\mathop {\lim }\limits_{x \to  - \infty } arctgx =  - \frac{\pi }{2}$

$\mathop {\lim }\limits_{x \to  + \infty } arctgx = \frac{\pi }{2}$

Хоризонтална асимптота:

\[\begin{array}{*{20}{c}}
\begin{gathered}
y = - \frac{\pi }{2} \hfill \\
x \to - \infty \hfill \\
\end{gathered} &\begin{gathered}
y = \frac{\pi }{2} \hfill \\
x \to + \infty \hfill \\
\end{gathered}
\end{array}\]

пр.6)   $f(x) = \frac{{2 - 2{x^2}}}{{{x^2} + 3}}$

$\mathop {\lim }\limits_{x \to  + \infty } \frac{{2 - 2{x^2}}}{{{x^2} + 3}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\frac{2}{{{x^2}}} - \frac{{2{x^2}}}{{{x^2}}}}}{{\frac{{{x^2}}}{{{x^2}}} + \frac{3}{{{x^2}}}}} = \frac{{0 - 2}}{{1 + 0}} =  - 2$

Десна хоризонтална асимптота: 

$y=-2$

$\mathop {\lim }\limits_{x \to  - \infty } \frac{{2 - 2{x^2}}}{{{x^2} + 3}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{2 - 2{{\left( { - t} \right)}^2}}}{{{{\left( { - t} \right)}^2} + 3}} = \mathop {\lim }\limits_{t \to  + \infty } \frac{{\frac{2}{{{t^2}}} - \frac{{2{t^2}}}{{{t^2}}}}}{{\frac{{{t^2}}}{{{t^2}}} + \frac{3}{{{t^2}}}}} = \frac{{0 - 2}}{{1 + 0}} =  - 2$

Лева хоризонтална асимптота: 

$y=-2$

пр.7)   $f(x) = \sqrt {{x^2} + 2}  - \sqrt {{x^2} - 5x}$

\[\begin{gathered}
Df:{x^2} + 2 \geqslant 0 \hfill \\
\forall x \in \mathbb{R} \hfill \\
\hfill \\
{x^2} - 5x \geqslant 0 \hfill \\
x\left( {x - 5} \right) \geqslant 0 \hfill \\
x = 0;x = 5 \hfill \\
\end{gathered} \]

$Df:x \in \left( { - \infty ;0} \right] \cup \left[ {5; + \infty } \right)$

\[\begin{gathered}
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 2} - \sqrt {{x^2} - 5x} } \right) \cdot \frac{{\sqrt {{x^2} + 2} + \sqrt {{x^2} - 5x} }}{{\sqrt {{x^2} + 2} + \sqrt {{x^2} - 5x} }} = \hfill \\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2} + 2 - \left( {{x^2} - 5x} \right)}}{{\sqrt {{x^2} + 2} + \sqrt {{x^2} - 5x} }} = \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2} + 2 - \left( {{x^2} - 5x} \right)}}{{\sqrt {{x^2} + 2} + \sqrt {{x^2} - 5x} }} = \hfill \\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{2}{x} + \frac{{5x}}{x} - \left( {{x^2} - 5x} \right)}}{{\frac{{\sqrt {{x^2} + 2} }}{{\sqrt {{x^2}} }} + \frac{{\sqrt {{x^2} - 5x} }}{{\sqrt {{x^2}} }}}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{0 + 5}}{{\sqrt 1 + \sqrt 1 }} = \frac{5}{2} \hfill \\
\end{gathered} \]

Десна хоризонтална асимптота: 

$y= \frac{5}{2}$

\[\begin{gathered}
\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 2} - \sqrt {{x^2} - 5x} } \right) = \mathop {\lim }\limits_{t \to + \infty } \left( {\sqrt {{{\left( { - t} \right)}^2} + 2} - \sqrt {{{\left( { - t} \right)}^2} - 5\left( { - t} \right)} } \right) = \hfill \\
x = - t \hfill \\
x \to - \infty \hfill \\
t \to + \infty \hfill \\
= \mathop {\lim }\limits_{t \to \infty } \left( {\sqrt {{t^2} + 2} - \sqrt {{t^2} + 5t} } \right) = \hfill \\ = \mathop {\lim }\limits_{t \to \infty } \left( {\sqrt {{t^2} + 2} - \sqrt {{t^2} + 5t} } \right) \cdot \frac{{\sqrt {{t^2} + 2} + \sqrt {{t^2} + 5t} }}{{\sqrt {{t^2} + 2} + \sqrt {{t^2} + 5t} }} = \hfill \\
= \mathop {\lim }\limits_{t \to \infty } \frac{{{t^2} + 2 - {t^2} - 5t}}{{\sqrt {{t^2} + 2} + \sqrt {{t^2} + 5t} }} = \mathop {\lim }\limits_{t \to \infty } \frac{{2 - 5t}}{{\sqrt {{t^2} + 2} + \sqrt {{t^2} + 5t} }} = \hfill \\
= \mathop {\lim }\limits_{t \to \infty } \frac{{\frac{2}{t} - 5}}{{\sqrt {1 + \frac{2}{{{t^2}}}} + \sqrt {1 + \frac{5}{t}} }} = - \frac{5}{2} \hfill \\
\end{gathered} \]

Лева хоризонтална асимптота: 

$y=- \frac{5}{2}$