Determine
$$\lim\limits_{x\to\infty}\frac{(x+7)^2\sqrt{x+2}}{7x^2\sqrt{x}-2x\sqrt{x}}.$$
Multiplying and dividing by $\sqrt{x}$ yields $$\frac{(x+7)^2\sqrt{x^2+2x}}{7x^3-2x^2}$$ where I would like to approximate the squareroot for sufficiently large $x$ with $$\frac{(x+7)^2\sqrt{x^2+2x+1}}{7x^3-2x^2}=\frac{(x+7)^2(x+1)}{7x^3-2x^2}=\frac{x^3(49/x^3+63/x^2+15/x+1)}{x^3(7-2/x)}\longrightarrow 1/7.$$
Can anyone confirm that my approximation is valid and does anyone know how to solve this in a more "usual" way like I did?