Find $\lim_{x\to \infty}\frac{\ln x}{x}=0$ without l'Hopital's rule 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!
 A: Notice that we have $\displaystyle 0 \leq \frac{\ln(x)}{x} \leq \frac{\sqrt{x}}{x} = \frac{1}{\sqrt{x}}$ for all $x \in \mathbb{R}^+$.
A: A different approach is to apply a substitution, that is $lnx=t$ 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.
A: 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.
A: 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
$$\lim_{x\to 0^+}x\ln x=0^-$$
