$ \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.

The mean value theorem gives $f(x+h) = f(x) + f'(x)h + {1 \over 2} f''(\xi) h^2$, where $\xi \in [x,x+h]$ (adjusted appropriately for sign of $h$).
This gives $|\sqrt{1+\theta} - (1+ {1 \over 2} \theta) | \le {1 \over 8} \theta^2$.
We have (if $x>0$),$\sqrt{x^2+2x} - \sqrt{x^2-7x} = x(\sqrt{1+{2 \over x}}-\sqrt{1-{7 \over 7}})$.
Then $| \sqrt{1+{2 \over x}}-\sqrt{1-{7 \over x}} - {1 \over 2}{9 \over x}| \le {1 \over 8} (({2 \over x})^2+({7 \over x})^2)$.
Multiplying through by $x$ and letting $x \to \infty$ yields the desired result.
Note: 
The point of this answer is that you can guess the answer by 
using $\sqrt{1+\theta} \approx 1+ {1 \over 2} \theta$ and
writing
$\sqrt{1+{2 \over x}}-\sqrt{1-{7 \over x}} \approx {1 \over 2}{9 \over x}$.
