Ex 1.
[tex]\lim _{x\to \:1}\frac{ln\left(x^2+x-1\right)}{ln\left(x^3+x-1\right)} => \lim _{x\to \:1}\left(\frac{\frac{2x+1}{x^2+x-1}}{\frac{3x^2+1}{x^3+x-1}}\right)=>\\lim _{x\to \:1}\left(\frac{\left(2x+1\right)\left(x^3+x-1\right)}{\left(x^2+x-1\right)\left(3x^2+1\right)}\right) =\frac{\left(2\cdot \:1+1\right)\left(1^3+1-1\right)}{\left(1^2+1-1\right)\left(3\cdot \:1^2+1\right)} = \boxed{\frac{3}{4}}\\[/tex]
(Aplicam regula lui L`Hospital)
Ex 2.
[tex]\lim _{x\to \:\:1}\frac{ln\left(1+x+x^2\right)+ln\left(1-x+x^2\right)}{x^2} = \frac{\ln \left(1+1+1^2\right)+\ln \left(1-1+1^2\right)}{1^2} =\frac{\ln \left(3\right)+\ln \left(1\right)}{1} =>\\\\ =\ln \left(3\right)+0 =\ln \left(3\right)[/tex]
