How do you find the limit of #s(n)=81/n^4[(n^2(n+1)^2)/4]# as #n->oo#?
To find the limit of ( s(n) = \frac{81}{n^4} \left( \frac{n^2(n+1)^2}{4} \right) ) as ( n ) approaches infinity (( n \to \infty )), you can simplify the expression first and then evaluate the limit:
- Expand ( \left( \frac{n^2(n+1)^2}{4} \right) ) to get ( \frac{n^4 + 2n^3 + n^2}{4} ).
- Multiply ( \frac{81}{n^4} ) by ( \frac{n^4 + 2n^3 + n^2}{4} ) to get ( \frac{81(n^4 + 2n^3 + n^2)}{4n^4} ).
- Simplify the expression to get ( \frac{81 + 162/n + 81/n^2}{4} ).
- As ( n ) approaches infinity, ( 162/n ) and ( 81/n^2 ) approach zero, leaving us with ( \frac{81}{4} ).
So, the limit of ( s(n) ) as ( n \to \infty ) is ( \frac{81}{4} ).
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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.
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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