<s>[INST] $ \lim\limits_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$ Find the limit: $$ \lim_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$$
I did the following:
\begin{align}
(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x}) = \frac{(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})}{1} \cdot \frac{(\sqrt{x^2 + 2x} + \sqrt{x^2 - 7x})}{(\sqrt{x^2 + 2x} + \sqrt{x^2 - 7x})}
\end{align}
I know the final answer is $\frac{9}{2}$. After multiplying by the conjugate, I see where the $9$ in the numerator comes from. I just can't remember how I solved the rest of the problem.
 [/INST] Expanding the rightmost term, we get:
$$
\frac{(x^2+2x) - (x^2-7x)}{\sqrt{x^2+2x} + \sqrt{x^2-7x}} = \frac{9x}{\sqrt{x^2+2x} + \sqrt{x^2-7x}}
$$
now just divide the numerator and denominator by $x = \sqrt{x^2}$:
$$
\frac{9}{\sqrt{1+\frac{2}{x}} + \sqrt{1-\frac{7}{x}}}.
$$
We have that $\sqrt{1+\frac{2}{x}} + \sqrt{1-\frac{7}{x}} \rightarrow 2$ as $x \rightarrow \infty$, so the final answer is $9/2$.
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