Răspuns:
Cred ca la prima se cere cotinuitatea in -1
Functia admite limite intr-un punct , daca limitele laterale sunt egale
Ls limita la stanga Ld limita la dreapta
1) Ls, x->-1 , x<-1 lim(x+5)=-1+5=4
Ld x->-1 x>-1 lim(3x²+1)=3(-1)²+1=3*1+1=4
Ls=Ld=4 functia are limita in x= -1
2)Ls, x<1 lim(x²-3x+2)=1²-3*1+2==1-3+2=0
Ld x->1 , x>1 lim lnx=ln1=0
Ls=Ld
3)Ls x->1 x<1 lim (x²+x-2)=1²+1-2=0
Ld x->1 x>1 lim(x+1)lnx=(1+1)ln1=2*0=0
Ls=Ld
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f(x)={3ˣ+1 x≤1
{ax+2 x>1
Ls x->1 x<1 lim (3ˣ+1)=3¹+1=4
Ld x->1 x>1 lim(ax+2)=a*1+2=a+2
Pui conditia Ls=Ld
a+2=4=>
a=2
Explicație pas cu pas:
