Why does $\frac{\log_{e}k}{k}$ approach $0$ as $k$ approaches $\infty$? 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?
First.

You can also understand it by graph.



let 

$$f(k)=k,\ g(k)=\ln k$$

then

$$f'(k)=1,\ g'(k)=\frac{1}{k}$$

If $k&gt;1$,

$$f'(k)&gt;g'(k),\ \lim_{k\to \infty}f'(k)=1&gt;\lim_{k\to \infty}g'(k)=0$$

notice that

$$f(1)&gt;g(1)$$

so

$$\lim_{k\to\infty}\frac{g(k)}{f(k)}=0$$

Second.

notice that

$$e=\lim_{x\to\infty}\left (1+\frac{1}{x}\right )^x$$

if $1/n$ is constant, we know

$$e=\lim_{x\to\infty}\left (1+\frac{n}{x}\right )^{\frac{x}{n}}$$

so

$$e^n=\lim_{x\to\infty}\left (1+\frac{n}{x}\right )^x$$

Important thing

$$e=\lim_{x\to\infty}\left (1+\frac{\bigstar}{x}\right )^\frac{x}{\bigstar}$$

$$\frac{\bigstar}{x}\times\frac{x}{\bigstar}=1$$
As $k\rightarrow 0^+$, $\ln(k)\rightarrow -\infty$ and $1/k\rightarrow +\infty$. So $\ln(k)/k\rightarrow-\infty$. The problem doesn't make sense for the two-sided limit since $\ln(x)$ is not defined for $x&lt;0$.