Wyznaczyć asymptoty funkcji

[wojtasekpl]

\(f(x) = \frac{2 \cdot 3^x+1}{3^x-9} \)

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

[eresh]

\(f(x) = \frac{2 \cdot 3^x+1}{3^x-9} \)

\(3^x\neq 3^2\\
x\neq 2\\
D=\mathbb{R}\setminus\{2\}\\
\Lim_{x\to 2}\frac{2\cdot 3^x+1}{3^x-9}=+\infty
\)
x=2 - asymptota pionowa

\(\Lim_{x\to \infty}\frac{2\cdot 3^x+1}{x(3^x-9)}=\Lim_{x\to\infty}\frac{2+\frac{1}{3^x}}{x(1-\frac{9}{3^x})}=0\\
\Lim_{x\to +\infty}\frac{2\cdot 3^x+1}{3^x-9}=\Lim_{x\to +\infty}\frac{2+\frac{1}{3^x}}{1-\frac{9}{3^x}}=2\\
\Lim_{x\to -\infty}\frac{2\cdot 3^x+1}{3^x-9}=-\frac{1}{9}\\
\)
asymptoty:
\(y=2\\
y=-\frac{1}{9}\\
x=2\)

[eresh]

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

\(D=\mathbb{R}\setminus\{0\}\\
\Lim_{x\to 0}\frac{3x^3+\cos x}{x^2}=\infty\)

\(\Lim_{x\to \infty}\frac{3x^3+\cos x}{x^3}=3\\
\Lim_{x\to \infty}(\frac{3x^3+\cos x}{x^2}-3x)=\Lim_{x\to \infty}\frac{3x^3+\cos x-3x^3}{x^2}=0\)
asymptoty:
\(y=3x\\
x=0\)

[wojtasekpl]

Dzięki serdeczne królowo.