evaluating some limits with $\ln(x)$ I don't understand how to prove these results.
$\lim\limits_{x \to +\infty}\dfrac{\ln{x}}{x} = 0$
$\lim\limits_{x \to 0^{+}}x\ln{x} = 0$
 A: Since $$\lim _{ x\rightarrow \infty  }{ \frac { x }{ { b }^{ x } } =0,b>1\quad  } $$ then for enough big $x$ : 
$$\frac { 1 }{ { b }^{ x } } <\frac { x }{ { b }^{ x } } <1$$
denote $b=e^{ \varepsilon }$ for small arbitrary $\varepsilon >0$ then we get:
$$\frac { 1 }{ { e }^{ \varepsilon x } } <\frac { x }{ e^{ \varepsilon x } } <1$$ or $$1<x<{ e }^{ \varepsilon x }$$ take the natural logarithm of both sides,we get $$\\ 0<\ln { x } <\varepsilon x$$ from this for enough big $x $  we finally get $$0<\frac { \ln { x }  }{ x } <\varepsilon $$ 
A: It might be useful to present an approach to evaluating these limits that forgoes use of L'Hospital's Rule.  Here, we begin by using the following definition of the exponential function $e^x$. 
$$e^x=\lim_{n\to \infty}\left(1+\frac xn\right)^n \tag 1$$
Then, it is straightforward to show from $(1)$ that for $x>0$
$$e^x\ge \left(1+\frac x2\right)^2 \tag 2$$
Inasmuch as $\log x$ is the inverse of $e^x$, then we have 
$$\lim_{x\to \infty}\frac{\log x}{x}=\lim_{x\to \infty}\frac{x}{e^x} \tag 3$$
From $(3)$ we have the inequality
$$0\le \frac{x}{e^x} \le \frac{x}{\left(1+\frac x2\right)^2} $$
which by the squeeze theorem implies that 
$$\lim_{x\to \infty}\frac{\log x}{x}=\lim_{x\to \infty} \frac{x}{e^x}=0$$
which was to be shown!

NOTE:
We remark that the limit $\lim_{x\to 0}x\log x$ can be shown to be equivalent to the limit $-\lim_{x\to \infty}\frac{\log x}{x}$ through the transformation $x\to 1/x$ and use of the identity $\log (1/x)=-\log x$.  Therefore, 
$$\lim_{x\to 0}x\log x=0$$
as was to be shown!
