Find $\lim_{x\to \infty}\frac{\ln x}{x}=0$ without l'Hopital's rule

Question

Can we find the limit $$\lim_{x\to \infty}\dfrac{\ln x}{x}=0$$ without using the l'Hopital's rule ? I did the change of variables $X=\ln x$ but it seems to be the same problem of finding the limit $\lim_{X\to 0^+}X\ln X$ for with I have no solution. Thank you for your help!

Answer 1

Notice that we have $\displaystyle 0 \leq \frac{\ln(x)}{x} \leq \frac{\sqrt{x}}{x} = \frac{1}{\sqrt{x}}$ for all $x \in \mathbb{R}^+$.

Answer 2

It is clear that $$\lim_{x\to\infty}\frac{\ln x}x= \lim_{x\to\infty}\frac{2\ln x}{x^2}. $$ On the other hand, $$0\leq\frac{2\ln x}{x^2}\leq \frac{2}x $$ since $x\geq\ln x.$ By squeeze theorem we conclude.

Answer 3

We can make it more clear by using FTC form of $\ln x=\int_{1}^{x}\frac{1}{x}dx$, and the inequality below which I'll prove later: $$\ln x=\int_{1}^{x}\frac{1}{t}dt\leq\int_{1}^{x}\frac{1}{\sqrt{t}}dt=2\sqrt{x}-2\tag{1}$$

To prove $(1)$, we only need to prove $\frac{1}{x}\leq\frac{1}{\sqrt{x}}$, given $x\geq 1$, and it is easy by squaring both sides then subtract $$ \begin{align} \frac{1}{x^2}&\leq\frac{1}{x}\\ \frac{x-1}{x^2}&\geq0\tag{2} \end{align} $$

$(2)$ is obviously true, now we can use $(1)$ and squeeze theorem to prove the statement $$ \begin{align} 0\leq&\ln x\leq 2\sqrt{x}-2\\ 0\leq&\frac{\ln x}{x}\leq\frac{2\sqrt{x}-2}{x}\\ 0\leq&\lim_{x\to\infty}\frac{\ln x}{x}\leq\lim_{x\to\infty}\frac{2\sqrt{x}-2}{x}=0\\ Q.E.D \end{align} $$

Similarly, we can substitute $x$ with $\frac{1}{u}$ to prove $$\lim_{x\to 0^+}x\ln x=0^-$$

Answer 4

A different approach is to apply a substitution, that is $\ln x=t$ so that the limit becomes : $\lim_{t\to \infty}\dfrac{t}{e^t}$ Now $e^t$ is a strictly increasing function (derivative?) and at $t=0$ the slope of the tangent is $1$ which is the same as the slope of the line $y=t$. So for all $t$ values more than $1$, the denominator grows faster than the numerator and thus this limit exists and is zero.

Answer 5

This result, and many others like it, can all be obtained from one simple limit: If $a>1,$ then $n/a^n \to 0$ as $n\to \infty$ through the natural numbers.

We don't need the differential calculus for this. Write $a = 1 + b,$ where $b>0.$ Then for $n\ge 2,$

$$a^n = (1+b)^n = 1 + n b + [n(n-1)/2]b^2 + \cdots \ge 1 + n b + [n(n-1)/2]b^2.$$

Why? The binomial theorem, that's why. Therefore

$$\frac{n}{a^n} \le \frac{n}{1 + n b + [n(n-1)/2]b^2}$$

for $n\ge 2.$ The limit of the right side is $0$ and we've proved the simple limit.

For the given limit, we can look at $(\ln e^n)/e^n = n/e^n \to 0$ from the simple limit. OK, we haven't taken care of the limit through all real values $x,$ but really, given the simple limit, you can take care of this part while filing your fingernails.

Answer 6

As an alternative, since by ratio-root criterion

$$\frac{a_{n+1}}{a_n} \rightarrow L\implies a_n^{\frac{1}{n}} \rightarrow L$$

with $a_n=n$ we have

$$\frac{n+1}{n} \rightarrow 1\implies n^{\frac{1}{n}} \rightarrow 1$$

and this implies as $x \to \infty$

$$x^{\frac1x} \to 1 \implies \frac{\log x}x\to 0$$

Answer 7

Let $f(x)=\dfrac{\ln x}{x}$

Notice for $x>e$ then $f(x)>0$ and $f'(x)=\frac{1-\ln(x)}{x^2}<0$

$f$ is decreasing and bounded inferiorly so it is bounded (by say $M$) in $[e,+\infty)$

$|f(x)|=\Big|\frac{2\ln(\sqrt{x})}{\sqrt{x}\sqrt{x}}\Big|=\underbrace{\Big|f(\sqrt{x})\Big|}_{<\,M}\times\Big|\frac 2{\sqrt{x}}\Big|\le \frac{2M}{\sqrt{x}}\to 0$

Answer 8

All that is needed to derive a more general result is $\ln(1+x) =\int_0^x \dfrac{dt}{1+t} $ for $x>0$.

Then, for any $0<c<1$,

$\begin{array}\\ \ln(1+x) &=\int_0^x \dfrac{dt}{1+t}\\ &<\int_0^x \dfrac{dt}{(1+t)^{1-c}}\\ &=\int_0^x (1+t)^{c-1}dt\\ &=\dfrac{(1+t)^c}{c}|_0^x\\ &=\dfrac{(1+x)^c-1}{c}\\ &\lt\dfrac{(1+x)^c}{c}\\ \text{or, for } x>1\\ \ln(x) &\lt\dfrac{x^c}{c}\\ \end{array} $

Setting $c=\frac12$ this is $\ln(x) \lt 2\sqrt{x} $.

More generally, setting $c=1/n$ this is

$\ln(x) \lt n x^{1/n} $.

This can also be used to show that, for $0 < a < 2$, $\dfrac{\ln(x)}{x^a} \to 0$ as $x \to \infty$.

Setting $c=a/2, 0 < a < 2$, $\ln(x) \lt\dfrac{x^{a/2}}{a/2} =\dfrac{2x^{a/2}}{a} $ so, dividing by $x^a$, $\dfrac{\ln(x)}{x^a} \lt\dfrac{2}{ax^{a/2}} \to 0$ as $x \to \infty$.