TITLE: Why does $\frac{\log_{e}k}{k}$ approach $0$ as $k$ approaches $\infty$?
QUESTION [5 upvotes]: A interesting problem: why is it that as $k$ approach infinity, $$\frac{\log_{e} k}{k}$$ approaches $0$?
Another problem: why $e^n=(1+n/x)^x$ as $x$ approach infinity?

REPLY [4 votes]: (For the first question.)
For any fixed $x > 0$, it follows from the binomial formula
$$
(1 + x)^k = {k \choose 0}x^0  + {k \choose 1}x^1  + {k \choose 2}x^2  +  \cdots  + {k \choose k-1}x^{k - 1}  + {k \choose k}x^k 
$$
that 
$$
(1 + x)^k  > \frac{{k(k - 1)}}{2}x^2  > k,
$$
for all $k$ greater than some positive integer $K$. Hence also
$$
\ln (1 + x)^k  > \ln k \;\; \forall k > K
$$
and, in turn,
$$
k \ln(1+x) > \ln k \;\; \forall k > K.
$$
Thus,
$$
0 < \frac{{\ln k}}{k} < \ln (1 + x) \;\; \forall k > K.
$$
Noting that $\varepsilon := \ln(1+x)$ is an arbitrary positive number, by definition of limit we have
$$
\mathop {\lim }\limits_{k \to \infty } \frac{{\ln k}}{k} = 0.
$$
EDIT: 
Noting that 
$$
\ln n^{1/n}  = \frac{{\ln n}}{n}, 
$$
any proof of
$$
\mathop {\lim }\limits_{n \to \infty } n^{1/n}  = 1
$$
can serve as a proof of
$$
\mathop {\lim }\limits_{n \to \infty } \frac{{\ln n}}{n} = 0.
$$
Here you can find an elegant proof of $\mathop {\lim }\nolimits_{n \to \infty } n^{1/n}  = 1$, using the AM-GM inequality.