<s>[INST] $\lim\limits_{x\to\infty}\frac{(x+7)^2\sqrt{x+2}}{7x^2\sqrt{x}-2x\sqrt{x}}$ 
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?
 [/INST] Hint: A standard thing to do is to divide top and bottom by $x^2\sqrt{x}$. 
The new top is
$$\left(1+\frac{7}{x}\right)^2 \sqrt{1+\frac{2}{x}},$$ 
and its behaviour for large $x$ is clear.
The new bottom is $7$ plus something tiny.
Remark: In the solution given in the OP, $\sqrt{x^2+2x}$ was replaced by $x+1$. True, this is fine, the change is indeed small. But if we are doing things formally, the replacement leaves a gap in the argument. 
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