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# Determine whether the sequence converges or diverges. If it converges, find the limit.$a_n = \frac {3 \sqrt {n}}{\sqrt {n} + 2}$

## The sequence converges to 3

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We're looking at limiters and goes to infinity of AM and for this limit, we can look at the denominator. Think about the term that's growing the fastest of this squared of in term. And then we divide the top in the bottom by that squared of end. So we have three on top and then one plus two over squared of end in the denominator. As N goes to infinity, this term is going to go to zero. So we just get three divided by one plus zero, which is just three. We converge, converge two three.

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