Question

Calcule o limite de​

Answer 1

Resposta:

(a-1)^5=-1 + 5 a - 10 a^2 + 10 a^3 - 5 a^4 + a^5

a^5-1 =(a-1)^5 - ( 5 a - 10 a^2 + 10 a^3 - 5 a^4 )

a^5-1 =(a-1)*(a^4+a³+a²+a+1)

Fazendo a=x^(1/5) , ficamos com

x-1 =(x^(1/5)-1)*(x^(4/5)+x^(3/5)+x^(2/5)+x^(1/5)+1)

x^(1/5)-1 = (x-1)/(x^(4/5)+x^(3/5)+x^(2/5)+x^(1/5)+1)

lim  [x^(1/5)  -1]/(x-1)

x-->1

lim  [ (x-1)/(x^(4/5)+x^(3/5)+x^(2/5)+x^(1/5)+1)]/(x-1)

x-->1

lim  1/[x^(4/5)+x^(3/5)+x^(2/5)+x^(1/5)+1]

x-->1

lim  [ 1/(1+1+1+1+1)]    =  1/5

x-->1

Answer 2

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$\displaystyle\sf\lim_{x \to 1}\dfrac{\sqrt[\sf5]{\sf x}-1}{x-1}\\\rm fac_{\!\!,}a~ u=\sqrt[\sf5]{\sf x}\implies x=u^5\\\sf u\to 1~quando~ x\to 1\\\displaystyle\sf\lim_{u \to 1}\dfrac{u-1}{u^5-1}\\\sf u^5-1=(u-1)\cdot(u^4+u^3+u^2+u+1)$

$\displaystyle\sf\lim_{ u \to 1}\dfrac{\diagup\!\!\!\!\!(u-\diagup\!\!\!\!\!1)}{\diagup\!\!\!\!\!(u-\diagup\!\!\!\!\!1)\cdot(u^4+u^3+u^2+u+1)}\\\displaystyle\sf\lim_{ u \to 1}\dfrac{1}{u^4+u^3+u^2+u+1}=\dfrac{1}{1^4+1^3+1^2+1+1}=\dfrac{1}{5}\checkmark$