lim(x²+7x+6)/x+1
x->-1
in acest moment avem nedeterminare 0/0
x²+7x+6=0
∆=49-24=25
x1=-7+5/2=-2/2=-1
x2=-7-5/2=-12/2=-6
deci limita se va scrie
lim[(x+1)(x+6)]/(x+1)=lim(x+6)=5
x->-1. x->-1
y=x+2 asimptota oblica
de aici avem m=1
n=2
unde m=limf(x)/x= lim(x²+ax+6)/(x²+x)=
x->infinit. x->infinit
=1(gradele sunt egale)
n=lim(x²+ax+6)/(x+1)-x=
x->infinit
=lim(x²+ax+6-x²-x)/(x+1)=
x->infinit
=lim [x(a-1)+6]/(x+1)=a-1 (gradele sunt egale)
x->infinit
dar totodata n=2
deci a-1=2
a=3
asimptota orizontala:
lim(x²+ax+6)/(x+1)=
x->infinit
=lim [x²(1+a/x+6/x²)]/x(1+1/x)=infinit
x->infinit
cum nu ne da un numar finit inseamma ca functia nu admite asimptota orizontala spre infinit oricare ar fi a apartine lui |R
