What is $\lim_{n\to\infty}\frac{2^{\log(n^2)}}{\sqrt{n^3}}$?

Question

I have to find out if $$2^{\log(n^2)} = \Omega (\sqrt{n^3}).$$

Now I have to find out if there exists a positive constant $c$ and a natural number $n_0$ such that $$2^{\log(n^2)} \ge c \sqrt{n^3},\quad \forall n > n_0.$$ Therefore I want to know if the limit of this function exists, because I couldn't find one. $$\lim_{n\to\infty}\frac{2^{\log(n^2)}}{\sqrt{n^3}}.$$

Answer 1

Take the logarithms and compare

$$\log2\log n^2=2\log 2\log n$$

and

$$\log\sqrt{n^3}=\frac32\log n.$$

Now,

$$2\log2<\frac32\iff 4\log 2<3\iff 16<e^3$$ because $16<\left(1+1+\dfrac12+\dfrac1{3!}\right)^3<e^3$.

Answer 2

Hint. Note that $$2^{\log(n^2)}=e^{\log(n^2)\log(2)}=n^{2\log(2)}.$$ Now compare it with $\sqrt{n^3}={n^{3/2}}$.

Answer 3

We have that

$$\frac{2^{\log(n^2)}}{\sqrt{n^3}}=\frac{4^{\log(n)}}{\sqrt{n^3}}=\frac{n^{\log(4)}}{\sqrt{n^3}}$$

and $\log 4 <\frac 3 2$.

Answer 4

As an alternative by $n^2 =2^x \to \infty$

$$\lim_{n\to\infty}\frac{2^{\log(n^2)}}{\sqrt{n^3}}=\lim_{x\to\infty}\frac{2^{\log(2^x)}}{2^{\frac34x}}=\lim_{x\to\infty}\frac{2^{x\log 2}}{2^{\frac34x}}$$

and $\log 2 <\frac 3 4$.