dimostrazioni di limiti tramite definizione in R

Teoremi

$\underset{x\to 0}{\lim }\sqrt{x}=0$

infatti $\forall \epsilon >0,\exists \delta >0:\forall x,$
$0<x<\delta \stackrel{?}{\Rightarrow }|\sqrt{x}|<\epsilon$
$\Updownarrow$
$0\le \sqrt{x}<\epsilon$
$\Updownarrow$
$0\le x<{\epsilon }^2$

basta prendere $\delta ={\epsilon }^2$

$x_0>0$, con $x_0\in \mathbb{R}$,
$\underset{x\to x_0}{\lim }\sqrt{x}=\sqrt{x_0}$

infatti $\forall \epsilon >0,\exists \delta >0:\forall x,$
$0<|x-x_0|<\delta \stackrel{?}{\Rightarrow }|\sqrt{x}-\sqrt{x_0}|<\epsilon$

$|\sqrt{x}-\sqrt{x_0}|=|(\sqrt{x}-\sqrt{x_0})\cdot \frac{\sqrt{x}+\sqrt{x_0}}{\sqrt{x}+\sqrt{x_0}}|=\frac{|x-x_0|}{\sqrt{x}+\sqrt{x_0}}\le \frac{|x-x_0|}{\sqrt{x}}$

quindi $|x-x_0|\le \epsilon \sqrt{x_0}$
$\delta =\epsilon \sqrt{x_0}$

Risorse