Question

Am nevoie urgent,multumesc din timp)

Answer 1

Răspuns:

Explicație pas cu pas:

1)  numaratorul are acelasi grad cu numitorul  ⇒lim F(x)x→∞=1/1=1

2) la fel  ⇒lim=1/2

3) x²+x-6=(x+3)(x-2)  si x²-3x+2=(x-2)(x-1)

lim (x+3)(x-2)/(x-2)(x-1)=lim (x+3)/(x-1) pr x→2  lim=5/1=5

4)lim(x²-1)/(2x²-x-1)=lim(x-1)(x+1)/(x²-x+x²-1)=lim(x+1)/(x+x+1) x→1=2/3

5)lim √(5x-1)/(3x-2) x→2=√(10-1)/(6-2)=3/4

6) lim (x²-3x+2)/(x+2)(x-2)[x+√(3x-2)] x→2=lim (x-1)/(x+2)[x+√(3x-2)]=

(2-1)/(2+1)[2+√(6-2)]=1/3·4=1/12