If the degree of numerator and denominator are equal, then limit will be leading coefficient of numerator divided by the
leading coefficient of denominator.
So then the limit would be 3/1 = 3.
Alternatively,
[tex]\displaystyle \lim_{x\to\infty}\dfrac{3x^2+6}{x^2-4}=\displaystyle \lim_{x\to\infty}\dfrac{3x^2+6}{x^2-4}\cdot\dfrac{1/x^2}{1/x^2}=\lim_{x\to\infty}\dfrac{3+\frac6{x^2}}{1-\frac4{x^2}} = \dfrac{3+0}{1-0}=\boxed{3}[/tex]
Hope this helps.
