There is a concept of **Rates of Growth** in Calculus.

Definition :- f(x) <<< g(x) as x $\rightarrow$ $\infty$ means growth rate of f(x) is very slow as compared to g(x) when x $\rightarrow$ $\infty$ ie. $\frac{f(x)}{g(x)} \rightarrow 0 \, \, as \, \, x\rightarrow \infty$

It is true when f and g are non-negative.

So, we can say that :-

1) $if \lim_{x\rightarrow \infty } \frac{f(x)}{g(x)} = \infty \, \, then\,\, f(x) >>> g(x)$ ie. growth rate of f(x) is higher than g(x)

(or)

2) $if \lim_{x\rightarrow \infty } \frac{f(x)}{g(x)} = 0 \, \, then \,\,g(x) >>> f(x)$ ie. growth rate of g(x) is higher than f(x)

Based on this fact we can say that :-

Rates of Growth of the following functions is :-

lnx << x^{p} << e^{x} << $e^{x^{2}}$ for p>0

and Rates of Decay should be :-

$\frac{1}{lnx}$ >> $\frac{1}{x^{p}}$ >> e^{-x} >> $e^{-x^{2}}$ , p>0

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Now , Rates of growth of Running Time of algorithms is same as order of growth of running time of algorithms. It is already mentioned in Cormen

According to Cormen ,

f(n) is asymptotically larger than g(n) if f(n) = $\omega$ (g(n))

(or)

$if \lim_{n\rightarrow \infty } \frac{f(n)}{g(n)} = \infty \, \, exist \, then\,\, f(n) \,\,becomes\,\, arbitrary \,\, large \,\, as\,\, compared\,\, to\,\, g(n) \,\,as\,\, n \,\,tends \,\,to\,\, infinity $

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Now , In this Question ,

On comparing f_{2} and f_{3 }

$\lim_{n \rightarrow \infty } \frac{n^{\frac{3}{2}}}{nlogn} = \lim_{n \rightarrow \infty } \frac{n^{\frac{1}{2}}}{logn} , when \,n \neq \infty \,\,$$Now , using \,L'H\hat{o}pital \, Rule , \lim_{n \rightarrow \infty } \frac{n^{\frac{1}{2}}}{logn} = \infty$

I have taken natural log above because it does not matter and we can convert one base to another and then solve. It will give same answer

So, f_{2} >>> f_{3}

Now , On comparing f_{1} and f_{4} :-

$\lim_{n \rightarrow \infty } \frac{2^{n}}{n^{logn}}$

Since , 2^{n} = (e^{ln2})^{n} = e^{n*ln2}

and n^{lnn }= (e^{ln n})^{ln n} = $e^{(ln n)^{2}}$

Since , exponential is an increasing function. So, comparing 2^{n} and n^{ln n} is same as comparing n*ln2 and (ln n)^{2}

So, By using $L'H\hat{o}pital's \, Rule $

$\lim_{n\rightarrow \infty } \frac{n*ln2}{(ln n)^{2}} = \infty$

So, n*ln2 > (ln n)^{2}

and $\lim_{n \rightarrow \infty } \frac{2^{n}}{n^{logn}}$ = $\infty$

So, now we can say f_{1} >>>f_{4}

Now, these 2 comparison of functions are enough to eliminate options in this question.