`lim_(xrarr oo) (n^p sin^2(n!))/(n+1),0ltplt1`, is equal to

1.	`lim_(xrarr oo) (n^p sin^2(n!))/(n+1),0ltplt1`, is equal to
Answer» Correct Answer - A We have , `lim_(xto oo) (n^p sin^2(n!))/(n+1)=lim_(ntooo)(sin^2(n!))/(1+(1)/(n))xxn^(p-1)` `lim_(xto oo) (n^p sin^2(n!))/(n+1)=lim_(ntooo)(sin^2(n!))/(n^(1-p))xx(1)/(1+(1)/(n))` ` rArr lim_(xto oo) (n^p sin^2(n!))/(n+1)=("An oscillating number")/(oo) xx1=0xx1=0`.

Discussion

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