find limit by rationalizing? I'm extremely new to calculus so please excuse my lack of lingo/formatting.!
Here is the problem:
$$\lim_{n\to\infty} \sqrt{5n^2 + 4n +2} - \sqrt{5n^2 - 2n - 1}$$
 A: 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 $$\frac{6}{2\sqrt 5} = \frac 3{\sqrt 5} = \frac {3\sqrt 5}{5}$$
A: Rationalizing expresion(multypling  nominator and denominator with ${ \sqrt{5n^2 + 4n +2} +\sqrt{5n^2 - 2n - 1}}$) you have $\lim_n\frac{6n+3}{ \sqrt{5n^2 + 4n +2} +\sqrt{5n^2 - 2n - 1}}$
And you have (dividing nominator and deonominator with n) that limit is 6/($\sqrt{5}+\sqrt{5}$) and that is 3/$\sqrt{5}$.
A: Set $\dfrac1n=h\implies h\to0^+$  as $n\to\infty$
$5n^2+4n+2=\dfrac{5+4h+2h^2}{h^2}$
$\sqrt{5n^2+4n+2}=\dfrac{\sqrt{5+4h+2h^2}}{\sqrt{h^2}}=\dfrac{\sqrt{5+4h+2h^2}}{+h}$ as $h>0$
So, we have $$\lim_{h\to0^+}\dfrac{\sqrt{5+4h+h^2}-\sqrt{5-2h-h^2}}h$$
$$=\lim_{h\to0^+}\dfrac{(5+4h+h^2)-(5-2h-h^2)}h\cdot\dfrac1{\lim_{h\to0^+}(\sqrt{5+4h+h^2}+\sqrt{5-2h-h^2})}$$
Hope the rest can be handled easily
