<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] $\bf{My\; Solution::}$ Given $\displaystyle \lim_{x\rightarrow \infty}\left(\sqrt{x^2+2x}-\sqrt{x^2-7x}\right) = \lim_{x\rightarrow \infty}x\left\{\left(1+\frac{2}{x}\right)^{\frac{1}{2}}-\left(1-\frac{7}{x}\right)^{\frac{1}{2}}\right\}$
Now Using Binomial Expansion::
$\displaystyle \lim_{x\rightarrow \infty}\left\{\left(1+\frac{1}{2}\cdot \frac{2}{x}+\frac{1}{2}\cdot -\frac{1}{2}\cdot \frac{4}{2x^2}......+\infty\right)-\left(1-\frac{1}{2}\cdot \frac{7}{x}-\frac{1}{2}\cdot -\frac{1}{2}\cdot \frac{49}{2x^2}.+\infty\right)\right\}$
So $\displaystyle \lim_{x\rightarrow \infty}\left(\sqrt{x^2+2x}-\sqrt{x^2-7x}\right) = 1+\frac{7}{2} = \frac{9}{2}$
</s>