# How do you find the limit of #s(n)=64/n^3[(n(n+1)(2n+1))/6]# as #n->oo#?

To find the limit of ( s(n) = \frac{64}{n^3} \left[ \frac{n(n+1)(2n+1)}{6} \right] ) as ( n \rightarrow \infty ), we can simplify the expression and then evaluate the limit. Simplifying the expression inside the brackets, we have:

[ \frac{n(n+1)(2n+1)}{6} = \frac{2n^3 + 3n^2 + n}{6} ]

Now, when ( n \rightarrow \infty ), the term with the highest degree dominates the expression. In this case, it's ( 2n^3 ). Dividing every term by ( n^3 ), we get:

[ \frac{2n^3 + 3n^2 + n}{6n^3} = \frac{2 + \frac{3}{n} + \frac{1}{n^2}}{6} ]

As ( n ) approaches infinity, ( \frac{3}{n} ) and ( \frac{1}{n^2} ) both approach zero. Thus, we're left with:

[ \lim_{n \to \infty} \frac{2 + 0 + 0}{6} = \frac{2}{6} = \frac{1}{3} ]

Therefore, the limit of ( s(n) ) as ( n \rightarrow \infty ) is ( \frac{1}{3} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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