Let y = √x, so the limit can be rewritten as
[tex]\displaystyle \lim_{y\to\infty}\left(\frac{\sqrt{y^6+y^2}}{y^2+3} - \frac{\sqrt{y^4+1}}{y+4}\right)[/tex]
Now,
[tex]\sqrt{y^6+y^2} = \sqrt{y^2\left(y^4+1\right)} = y\sqrt{y^4+1}[/tex]
so we can rewrite the limit further as
[tex]\displaystyle \lim_{y\to\infty}\left(\frac{y}{y^2+3} - \frac1{y+4}\right)\sqrt{y^4+1}[/tex]
Combine the rational terms:
[tex]\dfrac y{y^2+3} - \dfrac1{y+4} = \dfrac{y(y+4)-(y^2+3)}{(y^2+3)(y+4)} = \dfrac{4y-3}{(y^2+3)(y+4)}[/tex]
Then in the limit, we get
[tex]\displaystyle \lim_{y\to\infty}\frac{(4y-3)\sqrt{y^4+1}}{(y^2+3)(y+4)} = \lim_{y\to\infty}\frac{(4y^3-3y^2)\sqrt{1+\dfrac1{y^4}}}{y^3+4y^2+3y+12} \\\\ = \lim_{y\to\infty}\frac{\left(4-\dfrac3y\right)\sqrt{1+\dfrac1{y^4}}}{1+\dfrac4y+\dfrac3{y^2}+\dfrac{12}{y^3}} = \boxed{4}[/tex]
