Limit of $2^{(\log_2 n)^2}/2^{(\log_2 n)^3}$ I am trying to find the following limit 
$$\lim_{n \to \infty} \frac{2^{(\log_2 n)^2}} {2^{(\log_2 n)^3}}$$ 
I really don't know where to start and any help would be appreciated!
 A: Hint:
$$\frac{2^{\log_2(n)^2}}{2^{\log_2(n)^3}}=2^{\log_2(n)^2-\log_2(n)^3}$$
What is $\lim\limits_{n\to\infty}\left(\log_2(n)^2-\log_2(n)^3\right)$?
A: Hint
What is the limit of the logarithm of your expression?
A: We're going to apply two of the Laws of Logs. We will use 
\begin{eqnarray*}
\log_b(x^y) &=& y\log_b(x) \\ \\
\log_b\!\left(\frac{x}{y}\right) &=& \log_b(x)-\log_b(y)
\end{eqnarray*}
I'm going to take the log of your expression:
\begin{eqnarray*}
\log_2\left(\frac{2^{(\log_2 n)^2}}{2^{(\log_2 n)^3}}\right) &=&
\log_2\left(2^{(\log_2 n)^2}\right) - \log_2\left(2^{(\log_2 n)^3}\right) \\ \\
&=& (\log_2 n)^2\cdot\log_2(2) - (\log_2 n)^3\cdot\log_2(2) \\ \\
&=& (\log_2 n)^2 - (\log_2 n)^3 \\ \\
&=& (\log_2 n)^2 \cdot (1-\log_2n)
\end{eqnarray*}
Hopefully, you agree that $\log_2n \to \infty$ as $n \to \infty$. This means that 
$$\lim_{n \to \infty}(\log_2 n)^2 \cdot (1-\log_2n) = -\infty$$
Going back a few lines, tells us then that
$$\lim_{n \to \infty} \log_2\left(\frac{2^{(\log_2 n)^2}}{2^{(\log_2 n)^3}}\right) = -\infty$$
Looking at the graph $y = \log_2 x$ shows us that $\log_2 x \to -\infty$ as $x \to 0$ and hence
$$\lim_{n \to \infty} \frac{2^{(\log_2 n)^2}}{2^{(\log_2 n)^3}} = 0$$
A: Look at it this way:
$\log_2 n$
 gets large
for large $n$.
Therefore,
$(\log_2 n)^3$
gets larger than
$(\log_2 n)^2$.
Therefore
$2^{(\log_2 n)^3}$
gets much larger than
$2^{(\log_2 n)^2}$.
Therefore
$\frac{2^{(\log_2 n)^2}}{2^{(\log_2 n)^3}}
\to 0
$.
The rest is details.
A: After cancelling the numerator with the denominator what we get is 1/2^(log n to the base 2)
After that use the fact that 
X ^(log Y to the base X) is nothing but equal to Y itself.
So in your case it would simply be 1/n
And now if n is tending to infinity and is surely 0...
