Limita fundamaentala cu parametri

andrei Mar Noi 06, 2012 1:15 am · **andrei** Mesaje : 3 Reputatie : 0 Data de inscriere : 30/10/2012

Limita fundamaentala cu parametri

andrei Mar Noi 06, 2012 1:15 am

$\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{a{x^2} + bx}}{{{x^2} + x + 1}}} \right)^x} = {e^3},\,\,a,b \in \mathbb{R}{\rm{,}}\,a > 0.\,\,a,b = ?$

**Admin** Mier Noi 07, 2012 6:25 pm

Re: Limita fundamaentala cu parametri

Admin Mier Noi 07, 2012 6:25 pm

$\begin{array}{l} \mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{a{x^2} + bx}}{{{x^2} + x + 1}}} \right)^x} = {e^3} \Leftrightarrow \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{{a{x^2} + bx}}{{{x^2} + x + 1}} - 1} \right)^x} = \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{{{x^2}\left( {a - 1} \right) + x\left( {b - 1} \right) - 1}}{{{x^2} + x + 1}}} \right)^x}\\ \mathop {\lim }\limits_{x \to \infty } {\left[ {\underbrace {{{\left( {1 + \frac{{{x^2}\left( {a - 1} \right) + x\left( {b - 1} \right) - 1}}{{{x^2} + x + 1}}} \right)}^{\frac{{{x^2} + x + 1}}{{{x^2}\left( {a - 1} \right) + x\left( {b - 1} \right) - 1}}}}}_{ = e}} \right]^{\mathop {\lim }\limits_{x \to \infty } \frac{{x\left( {{x^2}\left( {a - 1} \right) + x\left( {b - 1} \right) - 1} \right)}}{{{x^2} + x + 1}}}}{ \Rightarrow ^{\mathop {\lim }\limits_{x \to \infty } \frac{{x\left( {{x^2}\left( {a - 1} \right) + x\left( {b - 1} \right) - 1} \right)}}{{{x^2} + x + 1}}}} = 3 \end{array}$
$\begin{array}{l} \mathop {\lim }\limits_{x \to \infty } \frac{{x\left( {{x^2}\left( {a - 1} \right) + x\left( {b - 1} \right) - 1} \right)}}{{{x^2} + x + 1}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^3}\left( {a - 1} \right) + {x^2}\left( {b - 1} \right) - x}}{{{x^2} + x + 1}} = 3 \Rightarrow \\ \left\{ \begin{array}{l} a - 1 = 0\\ b - 1 = 3 \end{array} \right. \Rightarrow a = 1,b = 4. \end{array}$