evaluating some limits with $\ln(x)$ I don't understand how to prove these results.
$\lim\limits_{x \to +\infty}\dfrac{\ln{x}}{x} = 0$
$\lim\limits_{x \to 0^{+}}x\ln{x} = 0$

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 $$0<\frac { \ln { x }  }{ x } <\varepsilon $$ 
