Question

what is the lim of x goes to infinity of x!/x^x and explain why

Answer 1

You could answer the question by considering the natural-number equivalent of the sequence \(\dfrac{n!}{n^n}\) and substituting successively large values of \(n\). You should notice that the denominator will get larger faster than the numerator, and so the limit must be 0.

Answer 2

In other words, you could apply a sort of ratio test to the expression. Compare the \(n\)th term of the sequence to the \((n+1)\)th term of the sequence. If the ratio of the two \(\left(\dfrac{a_{n+1}}{a_{n}}\right)\) is less than 1, then \(a_{n+1}<a_n\), which means the sequence is decreasing.

If you can show the sequence is also bounded in addition to the monotonicity, then it converges, which means the real-number equivalent must also converge to the same value.

Answer 3

Thanks!

Answer 4

yw