what is the limit of $(\log n)/n^{(1/100)}$ as n approaches infinity? I am kind of confused about this question, are we allowed to take l'Hopitals rule, can someone please show me the steps. 
 A: It is 0.
Yes, you can use  l'Hopitals rule, to find the limit of this function. Since numerator and Denominator each is limiting to infinity at infinity.


*

*Derivative( log(n) ) = 1/n

*Derivative( n1/100 ) = 1/100 * n - 99/100
So as per l'Hopitals rule : limn->inf [ ( log(n) ) / ( n1/100 ) ] = limn->inf [ 100 / ( n1/100 ) ] = 0.
Moreover, functions of the form xa always grow faster than logarithmic functions. Therefore, you can safely assume as limit turning out out be 0.
A: 
I think you simply need to take a extra logarythm :
$$
\lim_{n\to\infty} {{\log n} \over {n^{1 \over 100}}}
 = \lim_{n\to\infty} {e^{\log { {log n} \over {n^{1 \over 100}}}}} =
 \lim_{n\to\infty} e^{({ \log\log n - {1 \over 100} \log n})} = \lim_{x\to\infty} { e^{-x}} = 0
$$
Daniel
A: For sure, you can use L'Hospital rule.
Using algebra, consider the function $$f(n)=\frac{\log (n)}{n^{\frac 1 {100}}}$$ $$f'(n)=\frac{1}{n^{101/100}}-\frac{\log (n)}{100 n^{101/100}}$$ $$f''(n)=\frac{101 \log (n)}{10000 n^{201/100}}-\frac{51}{50 n^{201/100}}$$ The first derivative cancels if $n=e^{100}$ and $$f(e^{100})=\frac{100}{e}\qquad \text{and}\qquad f''(e^{100})=-\frac{1}{100 e^{201}}<0$$ So, the point  $n=e^{100}$ corresponds to a maximum and from there the function decreases in a monotonic manner to $0$.
If you set $n=e^{10^k}$, you will have $$f\left(e^{10^k} \right)=10^k e^{-10^{k-2}}$$ which goes very fast to $0$ as shown in the table below
$$\left(
\begin{array}{cc}
 1 & 9.04837 \\
 2 & 36.7879 \\
 3 & 0.0453999 \\
 4 &3.72 \times 10^{-40} \\
 5 & 5.08 \times 10^{-430} \\
 6 & 1.14 \times 10^{-4337} \\
 7 & 3.56 \times 10^{-43423} \\
 8 & 3.29 \times 10^{-434287}\\
 9 & 1.51 \times 10^{-4342936} 
\end{array}
\right)$$
A: $y:=\log(n) \rightarrow \exp(y) = n.$
$n^{1/100}= \exp(y/100)$.
$\dfrac{\log(n)}{n^{1/100}} =$
$\dfrac{y}{\exp(y/100)}=$
$\dfrac{100z}{\exp(z)}=: A(z)$, where 
$z:= (y/100)$.
$\rightarrow :$
$\lim_{z \rightarrow \infty} A(z) = 0.$
