Calcule o seguinte limite(Sem usar a L'Hospital) Resposta: 49/24.... Pergunta de ideia deRonnyMat

December 2019 1 54 Report

Calcule o seguinte limite(Sem usar a L'Hospital)

$\lim_{x \to 1} \frac{x^{100}-2x+1}{x^{50}-2x+1}$

Resposta: 49/24

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andresccp $\lim_no numeric noise key 1110 \fracno numeric noise key 1109-2x+1}no numeric noise key 1108-2x+1} \\\\$

briot ruffini:
$\boxed{1}\boxed{1_{x^{100}}}\boxed{0_{x^{99}}}\boxed{0_{x^{98}}}...\boxed{-2_x}\boxed{1}\\\\ \boxed{\;\;}\boxed{1_{x^{99}}}\boxed{1_{x^{98}}}\boxed{1_{x^{97}}}...\boxed{-1}\boxed{0}\\\\\ x^{100}-2x+1 = (x-1)(x^{99}+x^{98}+.... +x-1)}$

repetindo no denominador
$\boxed{1}\boxed{1_{x^{50}}}\boxed{0_{x^{49}}}\boxed{0_{x^{48}}}...\boxed{-2_x}\boxed{1}\\\\ \boxed{\;\;}\boxed{1_{x^{49}}}\boxed{1_{x^{48}}}\boxed{1_{x^{47}}}...\boxed{-1}\boxed{0}\\\\\ x^{50}-2x+1 = (x-1)(x^{49}+x^{48}+.... +x-1)}$

$\lim_{x \to 1} \frac{(x-1)(x^{99}+x^{98}+.... +x-1)}{(x-1)(x^{49}+x^{48}+.... +x-1)} \\\\ \lim_{x \to 1} \frac{x^{99}+x^{98}+.... +x-1}{x^{49}+x^{48}+.... +x-1} = \frac{ (1^{99}+1^{98}+1^{97}....+1-1)}{ (1^{49}+1^{48}+1^{47}....+1-1)} = \frac{99-1}{49-1} = \frac{49}{24}$

2 votes Thanks 2

RonnyMat vc 'e bom mano !!!

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