x→1 x<1 lim x^4/(x-1(x²+x+1)=1/3*lim x^4/(x-1)=1/3*1/(1-0-1)=1/-0=-∞
x→1, x>1 1/3 lim 1/(1+0-1)=1/3*1/+0=+∞
Ex 11 este gesit pt ca daca x=1 numitorul e 0 si functia ->∞, nu admite tangenta
Ex 10
f(1)=3+m+n)/2=2
m+n=1 formula 1
L=lim (1-1+f(x)/3)^x
=(1+(-3x^2-+3x^2+mx+n)/(3x^2+3)]^x
=lim(1+(mx+n)/(3x^2+3)]^x
Ridici paranteza dreapta la puterea (3x^2+3)/(mx+n)*(mx+n)/(3x^2+3)=1
L=lim [1+(mx+n)/(3x^2+3)]^[(3x^2+3)/(mx+n)]^x*(mx+n)*x/(3x²+3)]=e^lim(mx²+nx)/(3x²+3)
Pui conditia ca limita de la exponent sa fie 2.asta presupune ca m=6
inlocuiesti aceasta valoare in formula 1 si afli pe n
n=-5
