L=`underset(xtooo)lim((x^(2)+x-1)/(3x^(2)+2x+4))^((3x^(2)+x)/(x-2))`
`underset(xtooo)lim((x^(2)+x-1)/(3x^(2)+2x+4))^(underset(xtooo)lim(3x^(2)+x)/(x-2))`
Now, `underset(xtooo)lim(x^(2)+x-1)/(3x^(2)+2x+4)=underset(xtooo)lim(1+(1)/(x)-(1)/(x^(2)))/(3+(2)/(x)+(4)/(x^(2)))=1/3`
Also, `underset(xtooo)lim(3x^(2)+x)/(x-2)=underset(xtooo)lim(3x+(1)/(x))/(1-(2)/(x))tooo`
Thus, `L= ((1)/(3))^(oo)=0`