## Calculus: Early Transcendentals 8th Edition

The sequence converges to $\sqrt{4} = 2$.
We can take the limit inside of the square root. Since the highest degree of $n$ in the numerator is 2 and the highest degree in the denominator is 2, the limit as $n$ approaches infinity of the inner function will be that of $\frac{4n^2}{n^2} = 4$. Hence the sequence converges to $\sqrt{4} = 2$.