Question

1. qual o limite lim x -> 1 (1 - x ^ 5)/(x ^ 6 - 1) ?
2. lim x → 0 √ x + 2 − √ 2 / x ?

( principalmente a 1 e a 2 da imagem)

Answer 1

Explicação passo a passo:

Vamos lá!

1.

$\lim_{x \to 1} \frac{1-x^5}{x^6-1} \\\lim_{x\to 1}\frac{(1-x)\cdot(1^4 + 1^3x+1^2x^2+1x^3+x^4)}{(x^3-1)(x^3+1)}\\\lim_{x\to 1}\frac{(1-x)\cdot (1 +x +x^2 +x^3+x^4)}{[(x-1)(x^2+x+1)](x^3+1)}\\\lim_{x\to 1}\frac{-(x-1)\cdot (1 +x +x^2 +x^3+x^4)}{(x-1)\cdot(x^2+x+1)(x^3+1)}\\\lim_{x\to 1}\frac{- (1 +x +x^2 +x^3+x^4)}{(x^2+x+1)(x^3+1)}\\\lim_{x\to 1}\frac{-(1+1+1^2+1^3+1^3)}{(1^2+1+1)(1^3+1)} = \frac{-5}{3\cdot 2} = -\frac{5}{6}$

2.

$\lim_{x\to 0}\frac{\sqrt{x+2} - \sqrt{2}}{x}\\\lim_{x\to 0} \frac{\sqrt{x+2} - \sqrt{2}}{x}\cdot\frac{\sqrt{x+2} + \sqrt{2}}{\sqrt{x+2} + \sqrt{2}} \\\lim_{x\to 0}\frac{(\sqrt{x+2})^2 - (\sqrt{2})^2}{x\cdot(\sqrt{x+2} + \sqrt{2})}\\\lim_{x\to 0}\frac{x+2 -2}{x\cdot(\sqrt{x+2} + \sqrt{2})}\\\lim_{x\to 0}\frac{\backslash\!\!\!x}{\backslash\!\!\!x\cdot(\sqrt{x+2})+\sqrt{2}}\\\lim_{x\to 0}\frac{1}{(\sqrt{x+2})+\sqrt{2}} = \frac{1}{\sqrt{0+2} +\sqrt{2}} = \frac{1}{2\sqrt{2}}.$

Bons estudos!