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

Question

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?

Answer 1

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 $k>1$,

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

notice that

$$f(1)>g(1)$$

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

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

$$\frac{\bigstar}{x}\times\frac{x}{\bigstar}=1$$

Answer 2

(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.

Answer 3

1. Application of l'Hôpital's rule $$\lim_{k\rightarrow \infty }\frac{\log k}{k}=\frac{\lim_{k\rightarrow \infty }\left( \frac{d}{dk}\log k\right) }{\lim_{k\rightarrow \infty }\left( \frac{% dk}{dk}\right) }=\frac{\lim_{k\rightarrow \infty }\frac{1}{k}}{% \lim_{k\rightarrow \infty }1}=\frac{0}{1}=0.$$ Alternatively, if one changes variables, using the substitution recommended by Theo Buehler $k=e^l\rightarrow\infty $ as $l$ tends to $\infty$, the limit can be evaluated observing that for $k\ge 1$, $e^l\ge\dfrac{l^2}{2}$, one has $$0\le\frac{\log k}{k}=\frac{l}{e^l}\le\frac{2l}{l^2}=\frac{2}{l}.$$ Applying limits one gets by the squeeze theorem $\lim_{k\rightarrow \infty }\frac{\log k}{k}=0$. However, I am not able to show the inequality without the Taylor series for the exponential (see Steven Stadnicki's comment).

2. By definition of $e=\lim_{k\rightarrow \infty }\left( 1+\frac{1}{k}% \right) ^{k}$

$$\begin{eqnarray*} \lim_{x\rightarrow \infty }\left( 1+\frac{n}{x}\right) ^{x} &=&\lim_{x\rightarrow \infty }\left( 1+\frac{1}{\frac{x}{n}}\right) ^{x} \\ &=&\lim_{x\rightarrow \infty }\left( \left( 1+\frac{1}{\frac{x}{n}}\right) ^{% \frac{x}{n}}\right) ^{n} \\ &=&\left( \lim_{x\rightarrow \infty }\left( 1+\frac{1}{\frac{x}{n}}\right) ^{% \frac{x}{n}}\right) ^{n} \\ &=&(e)^{n}=e^{n}. \end{eqnarray*}$$

Answer 4

Another straightforward approach is through integrals. Since $x^{-1} \lt x^{t-1}$ for any $t\gt 0$ and $x\gt 1$, $\int_1^k x^{-1}\mathrm{d}x\lt \int_1^k x^{t-1}\mathrm{d}x$; $\log(k)\lt {k^t-1\over t}$. But choosing $t={1\over2}$ here gives $\log(k)\lt 2(k^{1/2}-1)$ and ${\log(k)\over k}\lt 2(k^{-1/2}-k^{-1})$, and the latter obviously goes to $0$ as $k\rightarrow\infty$. Note that you can easily adapt this to show that $\lim_{k\rightarrow\infty}(\log k/ k^\epsilon) = 0$ for any $\epsilon\gt 0$.

Answer 5

Since you get the indeterminate form $\infty / \infty$ as $k\to\infty$, you can use l'Hopital's rule:

$$\lim_{k\to\infty} \frac{\ln k}{k} = \lim_{k\to\infty} \frac{1/k}{1} = 0$$

Answer 6

Every time $k$ gets multiplied by $e$, $\log_e k$ increases by $1$. So $\frac{\log_e k}{k}$ is replaced by $$ \frac{1+\log_e k}{ek} $$ and this is less than half of $\frac{\log_e k}{k}$ when $k$ is bigger than....17, I think.

So every time $k$ gets multiplied by $e$, the expression gets cut down to less than half what it was before. Hence it must approach 0.

Answer 7

For the second question, first note that $(1+n/x)^x = \exp(x \log(1+n/x))$. Second, observe that the graph of the natural logarithm function $\log(x)$ is tangent to $x-1$ at $x=1$, so that $\log(1+\epsilon) = \epsilon + \mathcal O(\epsilon^2)$. Thus

$$\left(1+\frac n x\right)^x = \exp\left(x \log\left(1+\frac n x\right)\right) = \exp\left(x\left(\frac n x + \mathcal O\left(\frac 1 {x^2}\right)\right)\right) = \exp\left(n + \mathcal O\left(\frac 1 x\right)\right),$$

which tends to $\exp(n)$ as $x$ tends to $\infty$.

Not quite coincidentally, the approximation $\log(1+\epsilon) \approx \epsilon$ for $|\epsilon| \ll 1$ and the other techniques demonstrated above come handy in all sorts of calculations dealing with probabilities of unlikely events and their complements. It's also useful to note that the approximation gives a strict upper bound: $\log(1+\epsilon) \le \epsilon$, and the equality only holds when $\epsilon = 0 $.