Question

Calcule o seguinte Limite

$\displaystyle \lim_{x \to p} \frac{\sqrt[n]{x}-\sqrt[n]{p}}{x-p}$

Answer 1

De acordo com a premissa:

$\displaystyle\sqrt[n]{x} - \sqrt[n]{p} = \frac{x-p}{\displaystyle \sum\limits_{\displaystyle i=1}^{\displaystyle n} \, \, \, p^{\displaystyle \frac{i-1}{n}} \cdot \, \, \, x^{\displaystyle \frac{n-i}{n}} }$

Temos:

$\displaystyle \lim_{x \to p} \frac{ \sqrt[n]{x} - \sqrt[n]{p} }{x-p}= \\ \\ \\ \\ \\ \frac{\frac{\displaystyle x-p}{\displaystyle \sum\limits_{\displaystyle i=1}^{\displaystyle n} \, \, \, p^{\displaystyle \frac{i-1}{n}} \cdot \, \, \, x^{\displaystyle \frac{n-i}{n}} }}{x-p} = \\ \\ \\ \\ \\ \frac{\displaystyle (x-p)}{(\displaystyle \sum\limits_{\displaystyle i=1}^{\displaystyle n} \, \, \, p^{\displaystyle \frac{i-1}{n}} \cdot \, \, \, x^{\displaystyle \frac{n-i}{n}}) } \cdot \frac{1}{(x-p)}=$


$\displaystyle \frac{1}{\displaystyle \sum\limits_{\displaystyle i=1}^{\displaystyle n} \, \, \, p^{\displaystyle \frac{i-1}{n}} \cdot \, \, \, x^{\displaystyle \frac{n-i}{n}}} = \\ \\ \\ \\ \\ \frac{1}{\displaystyle \sum\limits_{\displaystyle i=1}^{\displaystyle n} \, \, \, p^{\displaystyle \frac{i-1}{n}} \cdot \, \, \, p^{\displaystyle \frac{n-i}{n}}} = \\ \\ \\ \\ \\ \frac{1}{ p^{\displaystyle \frac{i-1}{n}} \cdot \, \, \, p^{\displaystyle \frac{n-i}{n}}} =$


$\displaystyle \frac{1}{p^{\displaystyle \frac{i-1}{n}+\frac{n-i}{n}}} = \\ \\ \\ \\ \\ \frac{1}{p^{\displaystyle \frac{i-1+n-i}{n}}}= \\ \\ \\ \\ \\ \frac{1}{p^{\displaystyle \frac{n-1}{n}}}= \\ \\ \\ \\ \boxed{\boxed{\frac{1}{ n \, \sqrt[\displaystyle n]{p^{\displaystyle n-1}} } }}$