how can I solve the limes of $n*log_2(n)$ vs $n^{log_3(4)}$

Question

I want to manually calculate which of these two functions ($n*log_2(n)$ vs. $n^{log_3(4)}$) has a higher asymptotic increasing without using a calculator or any software.

My approach till now was:

lim n-> inf: \frac {$n*log:2(n)$} {$n^{log_3(4)}$}

Now use L´Hospital and derive each function: \frac {$log_2(n)$ + $1/ln(2)n$ } {$log_3(4) n^{log_3(4) -1}}

Now use L´Hospital again: \frac {$1/(ln(2)*n)$ + $1/(ln(2)*n) $} {$1/ln(3)4 $ * $n^{log_3(4)-1}$ + $log_3(4)-1 * n^{log_3(4)-2} * log_3(4) $}

My problem: If I calculate like that it results to a wrong solution. Does anyone have an idea how to solve that correctly?

Answer 1

Edit: I also noticed that your first derivative was incorrect.

Your first derivative and second evaluation of L' Hopitals rule is incorrect.

You start with:
f(n)=n*log2(n)
g(n)=n^(log3(4))

This gives:
f'(n)=log2(n) + n * (1/ln(2)) * n^(-1)
=log2(n) + 1/ln(2)
g'(n)=log3(4) * n^(log3(4)-1)

This gives:
f''(n)=(1/ln(2)) * n^(-1)
g''(n)=log3(4) * (log3(4)-1) * n^(log3(4)-2)

With your error in the first derivate you would have gotten f''(n)=(1/ln(2)) * n^(-1) - (1/ln(2)) * n^(-2), which still allows you to factor out n and results in the same final result.

Now that you have n in all of it, you can factor that out:
f''(n)/g''(n) = 1/[ln(2) * log3(4) * (log3(4)-1) * n^(log3(4)-2+1)]
= 1/[ln(2) * log3(4) * (log3(4)-1)] * n^(1-log3(4))]
Which now can be represented as:
k * n^(1-log3(4)) where k>0.
And the limit as this approaches infinity is 0. That means n^log3(4) has a greater asymptote than n * log2(n).

Alternatively, you can simplify first. Note that both have a factor of n which can be removed, so instead you can have:
f(n)=log2(n)
g(n)=n^(log3(4)-1)

f'(n)=(1/ln(2)) * n^(-1)
g'(n)=(log3(4)-1) * n^(log3(4)-2)

f'(n)/g'(n) = (1/ln(2)) * n^(-1-log3(4)+2)/(log3(4)-1)
=(1/ln(2)) * n^(1-log3(4))/(log3(4)-1)

Again, the limit is 0, meaning that n^(log3(4)) has a greater asymptote.

The only thing extra that is needed to know is that log3(4) is greater than 1 as 4 is greater than 3. That means (log3(4)-1)>0 and (1-log3(4))<0.

Also remember that the correct result may not be what you think it is. These 2 equations cross when n~= 30 000

Also, I'm not sure if this belongs here or on math.