I have a litle discution with a friend about the folowing limit:
limn→∞n√n!n
I would solve it like this: limn→∞n√n!nn=0
or
limn→∞n√nn√n−1⋯n√1n=1∗1∗1⋯∞=0
and in this 2º way would there be a problem with 1∞? I would say that no, because there is no functions involved, since as much as I know this undetermination is because you would whant to avoid the situation such as f(x)g(x) where f(x)→1 and g(x)→∞ Could anyone clarify this for me?
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