$\lim\limits_{x\to\infty}\frac{(x+7)^2\sqrt{x+2}}{7x^2\sqrt{x}-2x\sqrt{x}}$ 
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?

As $x \to +\infty$ we have $(x + 7)^2 \sim x^2$ (i.e. $\lim_{x \to \infty} \dfrac{(x+7)^2}{x^2} = 1$), $\sqrt{x+2} \sim \sqrt{x}$, and in the denominator 
$7 x^2 \sqrt{x} - 2 x \sqrt{x} \sim 7 x^2 \sqrt{x}$, so 
$$ \dfrac{(x+7)^2 \sqrt{x+2}}{7 x^2 \sqrt{x} - 2 x \sqrt{x}} \sim \dfrac{x^2 \sqrt{x}}{7 x^2 \sqrt{x}} = \frac{1}{7} $$
Thus the answer is $1/7$.
