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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?

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    @icurays1: I think I need a break... fixed this one, too.2012-11-12

2 Answers 2

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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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    @ChristianIvic$e$vic: If we are informally trying to find the limit, what you did is just fine. We do need to develop some intuition about what changes will make no difference. However, if this is a *formal* exercise in finding a limit, any replacement needs to be fully justified, by *showing* it makes no difference.2012-11-12
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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 $1/7$.