<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] Setting $\dfrac1x=h,$  the limit reduces to $$\lim_{h\to0^+}\frac{\sqrt{1+2h}-\sqrt{1-7h}}h$$
$$=\lim_{h\to0^+}\frac{1+2h-(1-7h)}{h(\sqrt{1+2h}+\sqrt{1-7h})}$$
$$=\lim_{h\to0^+}\frac9{\sqrt{1+2h}+\sqrt{1-7h}}$$
$$=\frac9{\sqrt{1+2\cdot0}+\sqrt{1-7\cdot0}}$$
</s>