Функције – асимптоте функција 3

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Одредити асимптоту графика функције:

пр.8)   $f(x) = \frac{{{x^3}}}{{{{(3 + x)}^2}}}$

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

$Df:{(3 + x)^2} \ne 0$

$3 + x \ne 0$

$x \ne  - 3$

$Df:\mathbb{R}\backslash \left\{ { - 3} \right\}$

B. A.:  

\[\begin{gathered}
\mathop {\lim }\limits_{x \to - {3^ - }} \frac{{{x^3}}}{{{{(3 + x)}^2}}} = \frac{{{{\left( { - {3^ - }} \right)}^3}}}{{{{\left( {3 + \left( { - {3^ - }} \right)} \right)}^2}}} = \frac{{ - 27}}{{{{\left( {{0^ - }} \right)}^2}}} = \frac{{ - 27}}{{{0^ + }}} = - \infty \hfill \\
\mathop {\lim }\limits_{x \to - 3 + } \frac{{{x^3}}}{{{{(3 + x)}^2}}} = \frac{{{{\left( { - {3^ + }} \right)}^3}}}{{{{\left( {3 + \left( { - {3^ + }} \right)} \right)}^2}}} = \frac{{ - 27}}{{{{\left( {{0^ + }} \right)}^2}}} = \frac{{ - 27}}{{{0^ + }}} = - \infty \hfill \\
\end{gathered} \]

B. A.:  $x =  - 3$

\[\begin{gathered}
\mathop {\lim }\limits_{x \to + \infty } \frac{{{x^3}}}{{{{(3 + x)}^2}}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{{{x^3}}}{{{x^3}}}}}{{\frac{{{{(3 + x)}^2}}}{{{x^2} \cdot x}}}} = \mathop {\lim }\limits_{x \to + \infty } \frac{1}{{{{(\frac{{3 + x}}{x})}^2} \cdot \frac{1}{x}}} = \hfill \\
= \mathop {\lim }\limits_{x \to + \infty } \frac{1}{{{{(\frac{3}{x} + 1)}^2} \cdot \frac{1}{x}}} = \frac{1}{{{{\left( {0 + 1} \right)}^2} \cdot 0}} = \frac{1}{0} = \infty \hfill \\
\end{gathered} \]

Нема десне Х. А.

\[\begin{gathered}
\mathop {\lim }\limits_{x \to - \infty } \frac{{{x^3}}}{{{{(3 + x)}^2}}} = \mathop {\lim }\limits_{t \to + \infty } \frac{{{{\left( { - t} \right)}^3}}}{{{{(3 + \left( { - t} \right))}^2}}} = \mathop {\lim }\limits_{t \to + \infty } \frac{{ - {t^3}}}{{{{\left( {3 - t} \right)}^2}}} = \hfill \\
x = - t \hfill \\
x \to - \infty \hfill \\
t \to + \infty \hfill \\
= \mathop {\lim }\limits_{t \to + \infty } \frac{{ - \frac{{{t^3}}}{{{t^3}}}}}{{{{\left( {\frac{3}{t} - 1} \right)}^2} \cdot \frac{1}{t}}} = \frac{{ - 1}}{0} = - \infty \hfill \\
\end{gathered} \]

Нема леве Х. А.

К. А.: десна $y = kx + n$

\[\begin{gathered}
k = \mathop {\lim }\limits_{x \to \infty } \frac{{f\left( x \right)}}{x} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{{x^3}}}{{{{\left( {3 + x} \right)}^2}}}}}{{\frac{x}{1}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}}}{{{{\left( {3 + x} \right)}^2}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}:{x^2}}}{{{{\left( {3 + x} \right)}^2}:{x^2}}} = \hfill \\
= \mathop {\lim }\limits_{x \to \infty } \frac{1}{{{{\left( {\frac{3}{x} + \frac{x}{x}} \right)}^2}}} = \frac{1}{{{1^2}}} = 1 \Rightarrow \underline {k = 1} \hfill \\
\hfill \\
n = \mathop {\lim }\limits_{x \to \infty } \left( {f\left( x \right) - kx} \right) = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^3}}}{{{{\left( {3 + x} \right)}^2}}} - x} \right) = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^3} - x{{\left( {3 + x} \right)}^2}}}{{{{\left( {3 + x} \right)}^2}}} = \hfill \\
= \mathop {\lim }\limits_{x \to \infty } \frac{{{x^3} - x\left( {9 + 6x + {x^2}} \right)}}{{{{\left( {3 + x} \right)}^2}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {{x^3} - 9x - 6{x^2} - {x^3}} \right):{x^2}}}{{{{\left( {3 + x} \right)}^2}:{x^2}}} = \hfill \\
= \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{ - 9x}}{{{x^2}}} - \frac{{6{x^2}}}{{{x^2}}}}}{{{{\left( {\frac{3}{{{x^2}}} + \frac{x}{{{x^2}}}} \right)}^2}}} = \frac{{ - 6}}{{{1^2}}} = - 6 \Rightarrow \underline {n = - 6} \hfill \\
\end{gathered} \]

ДЕСНА К. А.: $y = x - 6$

К. А.: лева $y = kx + n$

\[\begin{gathered}
k = \mathop {\lim }\limits_{x \to - \infty } \frac{{f\left( x \right)}}{x} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\frac{{{x^3}}}{{{{\left( {3 + x} \right)}^2}}}}}{{\frac{x}{1}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{{x^2}}}{{{{\left( {3 + x} \right)}^2}}} = \hfill \\
x = - t \hfill \\
x \to - \infty \hfill \\
t \to + \infty \hfill \\
= \mathop {\lim }\limits_{t \to \infty } \frac{{{t^2}}}{{{{\left( {3 + t} \right)}^2}}} = \mathop {\lim }\limits_{t \to \infty } \frac{{{t^2}:{t^2}}}{{{{\left( {3 + t} \right)}^2}:{t^2}}} = \mathop {\lim }\limits_{t \to \infty } \frac{1}{{{{\left( {\frac{3}{t} + \frac{t}{t}} \right)}^2}}} = \frac{1}{{{1^2}}} = 1 \Rightarrow \underline {k = 1} \hfill \\
\hfill \\
n = \mathop {\lim }\limits_{x \to - \infty } \left( {f\left( x \right) - kx} \right) = \mathop {\lim }\limits_{t \to \infty } \frac{{9t - 6{{\left( { - t} \right)}^2}}}{{{{\left( {3 - t} \right)}^2}}} = \mathop {\lim }\limits_{t \to \infty } \frac{{\left( {9t - 6{t^2}} \right):{t^2}}}{{{{\left( {3 - t} \right)}^2}:{t^2}}} = \hfill \\
= \mathop {\lim }\limits_{t \to \infty } \frac{{\frac{{9t}}{{{t^2}}} - \frac{{6{t^2}}}{{{t^2}}}}}{{{{\left( {\frac{3}{{{t^2}}} + \frac{t}{{{t^2}}}} \right)}^2}}} = \frac{{ - 6}}{{{{\left( { - 1} \right)}^2}}} = - 6 \Rightarrow \underline {n = - 6} \hfill \\
\end{gathered} \]

ЛЕВА К. А.: $y = x - 6$