[tex]\displaystyle\lim_{x\to\infty}\frac{3x^4+4}{(x^2-7)(4x^2-1)}[/tex]
Expanding the denominator gives
[tex](x^2-7)(4x^2-1)=4x^4-29x^2+7[/tex]
Then in the whole rational expression, we can divide through by [tex]x^4[/tex]. Since [tex]x\to\infty[/tex], we assume that [tex]x>0[/tex], so this is legal.
[tex]\dfrac{3x^4+4}{4x^4-29x^2+7}=\dfrac{3+\frac4{x^4}}{4-\dfrac{29}{x^2}+\dfrac7{x^4}}[/tex]
As [tex]x[/tex] gets arbitrarily large, the rational terms in the numerator and denominator become negligible. So for large [tex]x[/tex], we have
[tex]\dfrac{3x^4+4}{(x^2-7)(4x^2-1)}\approx\dfrac34[/tex]
And so the limit is [tex]\dfrac34[/tex].
