Zadanie 1.
[tex]f(x)=\frac{-2x^3-x^2+x-1}{x^2+3x+1}[/tex]
Asymptoty ukośnej szukamy w postaci [tex]y=ax+b[/tex].
Policzmy a:
[tex]a= \lim_{x \to \pm\infty} \frac{f(x)}{x}=\lim_{x \to \pm\infty} \frac{\frac{-2x^3-x^2+x-1}{x^2+3x+1}}{x}=\lim_{x \to \pm\infty} \frac{-2x^3-x^2+x-1}{x(x^2+3x+1)}=[/tex]
[tex]=\lim_{x \to \pm\infty} \frac{-2x^3-x^2+x-1}{x^3+3x^2+x}=\lim_{x \to \pm\infty} \frac{x^3(-2-\frac{1}{x}+\frac{1}{x^2}-\frac{1}{x^3})}{x^3(1+\frac{3}{x}+\frac{1}{x^2})}=\frac{-2}{1}=-2[/tex]
Policzmy b:
[tex]b=\lim_{x \to \pm\infty} (f(x)-ax)=\lim_{x \to \pm\infty} \left(\frac{-2x^3-x^2+x-1}{x^2+3x+1}+2x\right)=[/tex]
[tex]=\lim_{x \to \pm\infty} \left(\frac{-2x^3-x^2+x-1}{x^2+3x+1}+\frac{2x(x^2+3x+1)}{x^2+3x+1}\right)=\lim_{x \to \pm\infty} \left(\frac{-2x^3-x^2+x-1}{x^2+3x+1}+\frac{2x^3+6x^2+2x}{x^2+3x+1}\right)=\lim_{x \to \pm\infty} \frac{5x^2+3x-1}{x^2+3x+1}=\lim_{x \to \pm\infty} \frac{x^2(5+\frac{3}{x}-\frac{1}{x^2})}{x^2(1+\frac{3}{x}+\frac{1}{x^2})}=\frac{5}{1}=5[/tex]Ostatecznie asymptota ukośna to
[tex]y=-2x+5[/tex]
Zadanie 2.
Znajdźmy najpierw miejsca zerowe wyrażeń w liczniku i mianowniku.
[tex]2x^2-8x-10=0\ |:2\\x^2-4x-5=0\\\Delta=(-4)^2-4*1*(-5)=16+20=36\\\sqrt\Delta=6\\x_1=\frac{4-6}{2}=-1\\x_2=\frac{4+6}{2}=5\\\\-x^2-5x-4=0\ |*(-1)\\x^2+5x+4=0\\\Delta=5^2-4*1*4=25-16=9\\\sqrt\Delta=3\\x_1=\frac{-5-3}{2}=-4\\x_2=\frac{-5+3}{2}=-1[/tex]
Teraz policzmy granicę:
[tex]\lim_{x \to -1} \frac{2x^2-8x-10}{-x^2-5x-4}= \lim_{x \to -1} \frac{2(x+1)(x-5)}{-(x+1)(x+4)}=\lim_{x \to -1} \frac{2(x-5)}{-(x+4)}=\frac{2(-1-5)}{-(-1+4)}=\frac{-12}{-3}=4[/tex]
