[Limites] Limites pela definiçao

por **JoaoLuiz07** » Qui Ago 27, 2015 16:55

questao

por **adauto martins** » Sáb Ago 29, 2015 20:52

$\lim_{x\rightarrow 3}({x}^{2}-x-6)=0$ ...a questao eh...dado um $\epsilon=0.005\succ 0,\exists \delta \succ 0$ (q.temos q. encontrar) , $\left|(x-3) \right|\prec \delta$ eh tal q. $\left|({x}^{2}-x-6)-0 \right|\prec \epsilon=0.005$ ...entao...
$\left|{x}^{2}-x-6 \right|=\left|(x-2)(x+3) \right|\prec \left|x-2 \right|.\left|x+3 \right|\prec \0.005$
$\left|((x-3)+1)(x+3) \right|\prec (\left|x-3 \right|+1).\left|x+3 \right|\prec 0.005\Rightarrow (\delta +1).\left|x+3 \right|\prec 0.005$ $\Rightarrow \delta \prec (0.005/\left|x+3 \right|)-1...$ p/q. $\delta \succ 0\Rightarrow (0.005/(\left|x+3 \right|))-1\succ 0\Rightarrow \left|x+3 \right|\prec 0.005\Rightarrow -0.05\prec x+3 \prec 0.005\Rightarrow -3.005\prec x \prec -2.95$ ...como $\delta=min((0.005/\left|-3.05+3 \right|)-1),(0.005/\left|-2.95+3 \right|-1)$ ...logo podemos tomar $\delta =(0.005/\left|-2.95+3 \right|)-1\simeq 0$

[Limites] Limites pela definiçao

[Limites] Limites pela definiçao

[Limites] Limites pela definiçao

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