Pamiętajmy, że im większa wartość mianownika, tym liczba bardziej zmierza do zera, a więc:
x→∞limxn1=0; n≥1
a)
x→∞limx2−x+53x2+x−6=
=x→∞limx2(1−x1+x25)x2(3+x1−x26)=
=x→∞lim1−x1+x253+x1−x26=
=13=3
b)
x→−∞lim8x4+x2−x4x4−x3+4=
=x→−∞limx4(8+x21−x31)x4(4−x1+x44)=
=x→−∞lim8+x21−x314−x1+x44=
=84=21
c)
x→∞limx4−x+8(5−x)(x2−2)=
=x→∞limx4−x+85x2−10−x3+2x=
=x→∞limx4−x+8−x3+5x2+2x−10=
=x→∞limx3(x−x2x+x38)x3(−1+x5+x22−x310)=
=x→∞limx−x2x+x38−1+x5+x22−x310=
=∞−1= 0
d)
x→−∞lim(x+1)(x−x2)x4−2x2+4=
=x→−∞limx2−x3+x−x2x4−2x2+4=
=x→−∞lim−x3+xx4−2x2+4=
=x→−∞limx3(−1+x21)x3(x−x2+x34)=
=x→−∞lim−1+x21x−x2+x34=
=−1−∞=∞
e)
=x→∞lim(x2−4x3−x+2x2+1)=
=x→∞lim((x+2)(x−2)x3−x+2x2+1)=
=x→∞lim((x+2)(x−2)x3−(x−2)(x+2)(x2+1)(x−2))=
=x→∞lim(x+2)(x−2)x3−(x2+1)(x−2)=
=x→∞lim(x+2)(x−2)x3−(x3−2x2+x−2)=
=x→∞lim(x+2)(x−2)x3−x3+2x2−x+2=
=x→∞limx2−42x2−x+2=
=x→∞limx2(1−x24)x2(2−x1+x22)=
=12=2
f)
x→−∞lim(x2−xx3+1+x2+xx4)=
=x→−∞lim(x(x−1)x3+1+x(x+1)x4)=
=x→−∞lim(x(x−1)(x+1)(x3+1)(x+1)+x(x+1)(x−1)x4(x−1))=
=x→−∞limx(x+1)(x−1)(x3+1)(x+1)+x4(x−1)=
=x→−∞limx(x2−1)x4+x3+x+1+x5−x4=
=x→−∞limx3−xx5 +x3+x+1=
=x→−∞limx3(1−x21)x3(x2+1+x21+x31)=
=x→−∞lim1−x21x2+1+x21+x31=
=1∞=∞
g)
x→∞lim(2x5−x3+x−1)=
x→∞lim(x5(2−x21+x41−x51))=
=∞⋅(2−0+0−1)=∞
h)
x→−∞lim(x4+3x3−6)=
=x→−∞lim(x4(1+3x1−x46))=
=∞⋅(1+0−0)=∞
i)
x→−∞lim(−2x3+10x2−2)=
=x→−∞lim(x3(−2+x10−x32))=
=−∞⋅(−2−0+0)=∞
