Made easy test series

Question

Order the following: $\frac{e^{n \log n}}{n}, 2^{n \log n}, n^{\sqrt n}$

Comments

$\frac{e^{n logn}}{n} = \frac{n^{n}}{n} = n^{n-1}$

$n^{n^{1/2}}$

$2^{nlogn} = n^{nlog 2}$

Option C is correct

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how did you compare n^(n-1) and n^n*log2?

log2 will be less than 1 if we assume log base is e as you took for first complexity.

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@Magma

What should be the approach to solve such type of questions...I got stuck sometimes!

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@himgta check this out, http://bigocheatsheet.com