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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
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Best answer
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63 votes
$A < C$ and $A < D$ are straight forward.

$E < B$ and $C, D < E$ as $E$ and $B$ are exponential functions and $B$ having a larger base.

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.
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4 votes

Here we've to eliminate options one by one.

check 1st letter of every option,there we've a confusion about A & D, so now let's compare between A & D

$n^{1/3}$  &  $nlog^{9}n$   $\Rightarrow$   $\frac{logn}{3}$  & logn + 9loglogn

So clearly D is bigger than A, i.e option B is out.

Now let's take next two letter in the remaining  options d,c. here we've a confusion b/w d & c,

$nlog^{9}n$  &  $n^{7/4}$   $\Rightarrow$   $log^{9}n$  &  $n^{3/4}$  

take n = $10^{100}$  then  $n^{3/4}$  = $10^{75}$    &    $log^{9}n$ = $10^{18}$ i.e  D<C

Now we can eliminate option C & D also and remaining option A is the answer.

If we Compare B & E ,  $e^{n}$  & $1.0000001^{n}$  (take log base e on both side) 

It becomes n & n.$\ln(1.0000001)$  which clearly shows B>E

Finally a, d, c, e, b  which is option A

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