Please help to evaluate this limit: lim x->infinity (x^23/2^x)?
回答 (2)
2017-01-23 5:43 am
Consider the log. 23log(x) - x log(2). Clearly, log(x) is going to get much smaller than x. So the -x log(2) term is going to dominate. In the original expression, that means that 2^x is going to get much larger than x^23.
2017-01-23 5:36 am
lim x->infinity (x^23/2^x)=0
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