<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] $$\lim_{x\rightarrow \infty} (\sqrt{x^2+2x} - \sqrt{x^2-7x}) $$$$=\lim_{x\rightarrow \infty} \bigg(\sqrt{x^2+2x+1} - \sqrt{x^2-7x+\frac{49}{4}}\bigg) $$ $$=\lim_{x\rightarrow \infty} \bigg( x+1 - \bigg(x-\frac{7}{2}\bigg)\bigg) =\frac{9}{2} $$
</s>