O alta limita

florinelll Mar Apr 24, 2012 3:44 am

Cred ca se rezolva prin simplificare,dar nu reusesc...Ma puteti ajuta va rog ?

$\\ \lim_{x \to x_0}\frac{x^2-x-(x_0^2-x_0)}{x-x_0}$

**Admin** Mar Apr 24, 2012 7:21 am

Rezolvare :
$\begin{gathered} \mathop {\lim }\limits_{x \to {x_0}} \frac{{{x^2} - x - (x_0^2 - {x_0})}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{{x^2} - x_0^2}}{{x - {x_0}}} + \mathop {\lim }\limits_{x \to {x_0}} \frac{{ - x + {x_0}}}{{x - {x_0}}} \hfill \\ \left\{ \begin{gathered} \mathop {\lim }\limits_{x \to {x_0}} \frac{{{x^2} - x_0^2}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\left( {x - {x_0}} \right)\left( {x + {x_0}} \right)}}{{x - {x_0}}} = 2{x_0} \hfill \\ \mathop {\lim }\limits_{x \to {x_0}} \frac{{ - x - {x_0}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{ - \left( {x - {x_0}} \right)}}{{x - {x_0}}} = - 1 \hfill \\ \end{gathered} \right. \Rightarrow l = 2{x_0} - 2 \hfill \\ \end{gathered}$

O alta limita

O alta limita

Re: O alta limita