I tried using Stirling's approximation and d'Alambert's ratio test but can't get the limit. Could someone show how to evaluate this limit?
Answer
Use equivalents:
n√n!n∼∞(√2πn)1nn⋅ne=1e(2πn)12n
Now ln(2πn)12n=lnπ+ln2n2n→n→∞0, hence
n√n!n∼∞1e.
How to find lim without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
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