You should try to take $\displaystyle 4n^2$ out of the square root to find the limit of $\displaystyle \frac{2n}{\sqrt{4n^2+3}}$ when $\displaystyle n\to \infty$.
You should try to take $\displaystyle 4n^2$ out of the square root to find the limit of $\displaystyle \frac{2n}{\sqrt{4n^2+3}}$ when $\displaystyle n\to \infty$.
How do I take $\displaystyle 4n^2$ out of the square root?
You can factor $\displaystyle 4n^2$ : $\displaystyle \sqrt{4n^2+3}=\sqrt{4n^2\left(1+\frac{3}{4n^2}\rig ht)}
=\sqrt{4n^2}\sqrt{1+\frac{3}{4n^2}}=\ldots$
You can factor $\displaystyle 4n^2$ : $\displaystyle \sqrt{4n^2+3}=\sqrt{4n^2\left(1+\frac{3}{4n^2}\rig ht)}
=\sqrt{4n^2}\sqrt{1+\frac{3}{4n^2}}=\ldots$