Why the limit of $\frac{\log n}{n} \int_2^n \frac{1}{\log x} d x $ is $1$? $$
\lim_{n \rightarrow \infty} \frac{\log n}{n} \int_2^n \frac{1}{\log x}
   d x = 1
$$
I tried to find the integral but the integral by itself does not converge.
When applying the Mean Value Theorem I can find that one side of my inequality is equal to 1 but I cannot prove the equality.  
 A: By L'Hospital, 
$$\lim_{n \rightarrow \infty} \frac{\log n}{n} \int_2^n \frac{1}{\log x}
   d x = \lim_{n \rightarrow \infty} \frac{1/\log(n)}{\frac{d}{d n} \frac{n}{\log n}} = \lim_{n \rightarrow \infty} \frac{1/\log(n)}{1/\log(n) - 1/\log(n)^2} = \lim_{n \rightarrow \infty} \frac{1}{1- 1/\log(n)} = 1
$$
A: Integration by parts gives
$$
\begin{align}
&\int_2^n\frac{\mathrm{d}x}{\log(x)}\\
&=\left.\frac{x}{\log(x)}\right]_2^n+\int_2^n\frac{\mathrm{d}x}{\log(x)^2}\\
&=\frac{n}{\log(n)}-\frac{2}{\log(2)}+O\!\left(\frac{n}{\log(n)^2}\right)\tag{1}
\end{align}
$$
The big-O estimate is valid since for $n\ge e^3$,
$$
\begin{align}
\int_2^n\frac{\mathrm{d}x}{\log(x)^2}
&=\int_2^{e^3}\frac{\mathrm{d}x}{\log(x)^2}+\int_{e^3}^n\frac{\mathrm{d}x}{\log(x)^2}\\
&\le\int_2^{e^3}\frac{\mathrm{d}x}{\log(x)^2}
+3\int_{e^3}^n\left(\frac1{\log(x)^2}-\frac2{\log(x)^3}\right)\mathrm{d}x\\
&=\int_2^{e^3}\frac{\mathrm{d}x}{\log(x)^2}
+3\left[\frac{x}{\log(x)^2}\right]_{e^3}^n\\
&=\int_2^{e^3}\frac{\mathrm{d}x}{\log(x)^2}
-\frac{e^3}3+3\frac{n}{\log(n)^2}\tag{2}
\end{align}
$$
Multiply $(1)$ by $\frac{\log(n)}{n}$ and take the limit as $n\to\infty$.
