find limit by rationalizing? I'm extremely new to calculus so please excuse my lack of lingo/formatting.!
Here is the problem:
$$\lim_{n\to\infty} \sqrt{5n^2 + 4n +2} - \sqrt{5n^2 - 2n - 1}$$

To rationalize:


*

*Multiply by $$\frac{\sqrt{5n^2 + 4n +2} + \sqrt{5n^2 - 2n -
   1}}{\sqrt{5n^2 + 4n +2} + \sqrt{5n^2 - 2n - 1}}$$
$$\left(\sqrt{5n^2 + 4n +2} - \sqrt{5n^2 - 2n -
   1}\right)\cdot\frac{\sqrt{5n^2 + 4n +2} + \sqrt{5n^2 - 2n -
   1}}{\sqrt{5n^2 + 4n +2} + \sqrt{5n^2 - 2n - 1}}$$ $$= \frac{6n+3}{\sqrt{5n^2 + 4n +2} + \sqrt{5n^2 - 2n - 1}}$$

*Now, divide numerator and denominator by $n$.
You should end up with a limit $$\frac{6}{2\sqrt 5} = \frac 3{\sqrt 5} = \frac {3\sqrt 5}{5}$$
