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Evaluate lim(ntooo) (n^(p)sin^(2)(n!))/(...

Evaluate `lim_(ntooo) (n^(p)sin^(2)(n!))/(n+1),"where "0ltplt1.`

Text Solution

Verified by Experts

We have `underset(ntooo)lim(n^(p)sin^(2)(n!))/(n+1)" "((oo)/(oo)"from")`
`=underset(ntooo)lim(sin^(2)(n!))/(n^(1-p)(1+1/n))`
`=("some number between 0 and1")/(ooxx1)=0" "("as "0ltplt1)`
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