Limita - eliminarea nedeterminarii prin simplificare

florinelll Joi Feb 23, 2012 11:47 pm

Cum restrang numitorul ca sa pot duce la bun sfarsit exercitiul ? Am auzit ...cumva,schema lui Horner folosind tabel ? Link-uri catre ceva documentatie ? Pana acum nu am reusit sa inteleg..

$\lim_{x\rightarrow 1}\frac{x^3-3x+2}{x^4-4x^2+3}$

**Admin** Vin Feb 24, 2012 12:25 am

Rezolvarea expandata a problemei ar fi :
$\begin{array}{l} \frac{{{x^3} - 3x + 2}}{{{x^4} - 4{x^2} + 3}} = \frac{{{x^3} - 3x + 3 - 1}}{{{x^4} - 4{x^2} + 4 - 1}} = \frac{{{x^3} - 1 - 3\left( {x - 1} \right)}}{{{x^4} - 1 - 4\left( {{x^2} - 1} \right)}} = \frac{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right) - 3\left( {x - 1} \right)}}{{\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right) - 4\left( {{x^2} - 1} \right)}}\\ \frac{{\left( {x - 1} \right)\left( {{x^2} + x + 1 - 3} \right)}}{{\left( {{x^2} - 1} \right)\left( {{x^2} + 1 - 4} \right)}} = \frac{{{x^2} + x - 2}}{{\left( {x + 1} \right)\left( {{x^2} - 3} \right)}}\\ \mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - 3x + 2}}{{{x^4} - 4{x^2} + 3}} = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} + x - 2}}{{\left( {x + 1} \right)\left( {{x^2} - 3} \right)}} = 0 \end{array}$

florinelll Mar Feb 28, 2012 11:55 pm

Va multumesc ! Very Happy

Continut sponsorizat

Limita - eliminarea nedeterminarii prin simplificare

Limita - eliminarea nedeterminarii prin simplificare

Re: Limita - eliminarea nedeterminarii prin simplificare

Re: Limita - eliminarea nedeterminarii prin simplificare

Re: Limita - eliminarea nedeterminarii prin simplificare