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Arrange the following functions in increasing asymptotic order:

1. $n^{1/3}$
2. $e^n$
3. $n^{7/4}$
4. $n \log^9n$
5. $1.0000001^n$

1. a, d, c, e, b
2. d, a, c, e, b
3. a, c, d, e, b
4. a, c, d, b, e

A < C and A < D

E < B

and

C, D < E as E is exponential function.

Now, we just need to see if C or D is larger.

In C we have a term $n^{3/4}$ and correspondingly in D we have $\log^9n$ (after taking $n$ out).

$n^{3/4}$ is asymptotically larger than $\log^9n$ as when $n = 10^{100}, \log^9 n$ gives $100^9$, while $n^{3/4}$ gives $10^{75} > 100^{37}$ a much higher value and this is true for all higher values of $n$. So, D < C.

Thus A is correct.
edited by
$10^{75} = 10.{10^{2}}^{37}= 10.100^{37}$
@arjun sir for any large value of n value of option e is less than option c.please correct me where am i going wrong
not getting how E be in 4 position

i caluculated its value to large n

which still at 1.000..?
if you have doubt you can try plotting the graph and see. Large value means some large value - can be even one with 100 digits - for the same reason we cannot always find this by substituting values.
it is very confusing sir any other way of solving this question