![\lim_{x \rightarrow 0}\sqrt[n]x - \sqrt[n]p{}{} / x - p](/latexrender/pictures/a7f22620032ff7aceba3a45471e3768d.png "\lim_{x \rightarrow 0}\sqrt[n]x - \sqrt[n]p{}{} / x - p")
Alguém pode me ajudar??
por LuizAquino » Sáb Abr 07, 2012 16:06
Priscilla Correa escreveu:
Alguém pode me ajudar??
![\lim_{x \to 0}\sqrt[n]{x} - \frac{\sqrt[n]{p}}{x} - p](/latexrender/pictures/92e91a908a724eb347af1b8da1846918.png "\lim_{x \to 0}\sqrt[n]{x} - \frac{\sqrt[n]{p}}{x} - p")
![\lim_{x \to p}\frac{\sqrt[n]{x} - \sqrt[n]{p}}{x - p}](/latexrender/pictures/7d25615a280b19a531a71b9386e8a584.png "\lim_{x \to p}\frac{\sqrt[n]{x} - \sqrt[n]{p}}{x - p}")
![\lim_{x \to p}\left(\sqrt[n]{x} - \sqrt[n]{p}\right)/(x - p)](/latexrender/pictures/cc844929dacec3a3f78b5b548166555e.png "\lim_{x \to p}\left(\sqrt[n]{x} - \sqrt[n]{p}\right)/(x - p)")
![\lim_{x \to p}\frac{\sqrt[n]{x} - \sqrt[n]{p}}{x - p} = \lim_{x \to p} \frac{\sqrt[n]{x} - \sqrt[n]{p}}{\left(\sqrt[n]{x}\right)^n - \left(\sqrt[n]{p}\right)^n}](/latexrender/pictures/b4a589302a6689d509ca436dfb2a3a3e.png "\lim_{x \to p}\frac{\sqrt[n]{x} - \sqrt[n]{p}}{x - p} = \lim_{x \to p} \frac{\sqrt[n]{x} - \sqrt[n]{p}}{\left(\sqrt[n]{x}\right)^n - \left(\sqrt[n]{p}\right)^n}")
por Priscilla Correa » Sáb Abr 07, 2012 16:15
por LuizAquino » Sáb Abr 07, 2012 16:19
Priscilla Correa escreveu:Obrigada pela resposta, eu fiquei meio confusa na hora de escrever a função e acabei digitando errado.
Então, eu resolvi e deu 1/0 (um sobre zero). Será que é isso mesmo???
por Priscilla Correa » Sáb Abr 07, 2012 16:33
![\lim_{x \rightarrow p}(\sqrt[n]x - \sqrt[n]p{}{}) / (x - p)
= \lim_{x \rightarrow p}(\sqrt[n]x - \sqrt[n]p{}{})(\sqrt[n]x + \sqrt[n]p{}{})/ (x - p)(\sqrt[n]x + \sqrt[n]p{}{})= \lim_{x \rightarrow p} 1/(\sqrt[n]x + \sqrt[n]p{}{}) = 1/(\sqrt[n]p + \sqrt[n]p{}{})](/latexrender/pictures/424b363154efd31ac8399945d56983f2.png "\lim_{x \rightarrow p}(\sqrt[n]x - \sqrt[n]p{}{}) / (x - p)
= \lim_{x \rightarrow p}(\sqrt[n]x - \sqrt[n]p{}{})(\sqrt[n]x + \sqrt[n]p{}{})/ (x - p)(\sqrt[n]x + \sqrt[n]p{}{})= \lim_{x \rightarrow p} 1/(\sqrt[n]x + \sqrt[n]p{}{}) = 1/(\sqrt[n]p + \sqrt[n]p{}{})")
por LuizAquino » Sáb Abr 07, 2012 16:54
Priscilla Correa escreveu:Eu refiz e cheguei a outro resultado.
Será que está certo??
![\left(\sqrt[n]{x} - \sqrt[n]{p}\right)\right\left(\sqrt[n]{x} + \sqrt[n]{p}\right)](/latexrender/pictures/9b2492a0192f29c4465c28e77cea554a.png "\left(\sqrt[n]{x} - \sqrt[n]{p}\right)\right\left(\sqrt[n]{x} + \sqrt[n]{p}\right)")
é igual a x - p.![\left(\sqrt[3]{x} - \sqrt[3]{p}\right)\right\left(\sqrt[3]{x} + \sqrt[3]{p}\right) = \left(\sqrt[3]{x}\right)^2 - \left(\sqrt[3]{p}\right)^2 \neq x - p](/latexrender/pictures/661d5ced03d4a6f842f7ddeee08aa636.png "\left(\sqrt[3]{x} - \sqrt[3]{p}\right)\right\left(\sqrt[3]{x} + \sqrt[3]{p}\right) = \left(\sqrt[3]{x}\right)^2 - \left(\sqrt[3]{p}\right)^2 \neq x - p")
![= \lim_{x\to p}\frac{\sqrt[n]{x} - \sqrt[n]{p}}{(\sqrt[n]{x} - \sqrt[n]{p})(\sqrt[n]{x}^{n-1} + \sqrt[n]{x}^{n-2}\sqrt[n]{p} + \sqrt[n]{x}^{n-3}\sqrt[n]{p}^2+\ldots \sqrt[n]{x}^2\sqrt[n]{p}^{n-3} + \sqrt[n]{x}\sqrt[n]{p}^{n-2} + \sqrt[n]{p}^{n-1})}](/latexrender/pictures/c19e8794a432ec461382d19217519ac7.png "= \lim_{x\to p}\frac{\sqrt[n]{x} - \sqrt[n]{p}}{(\sqrt[n]{x} - \sqrt[n]{p})(\sqrt[n]{x}^{n-1} + \sqrt[n]{x}^{n-2}\sqrt[n]{p} + \sqrt[n]{x}^{n-3}\sqrt[n]{p}^2+\ldots \sqrt[n]{x}^2\sqrt[n]{p}^{n-3} + \sqrt[n]{x}\sqrt[n]{p}^{n-2} + \sqrt[n]{p}^{n-1})}")
![= \lim_{x\to p}\frac{1}{\sqrt[n]{x}^{n-1} + \sqrt[n]{x}^{n-2}\sqrt[n]{p} + \sqrt[n]{x}^{n-3}\sqrt[n]{p}^2+\ldots \sqrt[n]{x}^2\sqrt[n]{p}^{n-3} + \sqrt[n]{x}\sqrt[n]{p}^{n-2} + \sqrt[n]{p}^{n-1}}](/latexrender/pictures/ec764a5d42aa7595b872abd5c8c65756.png "= \lim_{x\to p}\frac{1}{\sqrt[n]{x}^{n-1} + \sqrt[n]{x}^{n-2}\sqrt[n]{p} + \sqrt[n]{x}^{n-3}\sqrt[n]{p}^2+\ldots \sqrt[n]{x}^2\sqrt[n]{p}^{n-3} + \sqrt[n]{x}\sqrt[n]{p}^{n-2} + \sqrt[n]{p}^{n-1}}")
![\lim_{x \to p}\frac{\sqrt[3]{x} - \sqrt[3]{p}}{x - p} = \lim_{x \to p} \frac{\sqrt[3]{x} - \sqrt[3]{p}}{\left(\sqrt[3]{x}\right)^3 - \left(\sqrt[3]{p}\right)^3}](/latexrender/pictures/920673a53c1c1a2ff76565584347eda7.png "\lim_{x \to p}\frac{\sqrt[3]{x} - \sqrt[3]{p}}{x - p} = \lim_{x \to p} \frac{\sqrt[3]{x} - \sqrt[3]{p}}{\left(\sqrt[3]{x}\right)^3 - \left(\sqrt[3]{p}\right)^3}")
![= \lim_{x\to p}\frac{\sqrt[3]{x} - \sqrt[3]{p}}{\left(\sqrt[3]{x} - \sqrt[3]{p}\right)\left(\sqrt[3]{x}^{2} + \sqrt[3]{x}\sqrt[3]{p} + \sqrt[3]{p}^2\right)}](/latexrender/pictures/ab2a22f401583e947560e206556b18bd.png "= \lim_{x\to p}\frac{\sqrt[3]{x} - \sqrt[3]{p}}{\left(\sqrt[3]{x} - \sqrt[3]{p}\right)\left(\sqrt[3]{x}^{2} + \sqrt[3]{x}\sqrt[3]{p} + \sqrt[3]{p}^2\right)}")
![= \lim_{x\to p}\frac{1}{\sqrt[3]{x}^{2} + \sqrt[3]{x}\sqrt[3]{p} + \sqrt[3]{p}^2}](/latexrender/pictures/dc482690dafb549a9993a5342708e544.png "= \lim_{x\to p}\frac{1}{\sqrt[3]{x}^{2} + \sqrt[3]{x}\sqrt[3]{p} + \sqrt[3]{p}^2}")
Voltar para Cálculo: Limites, Derivadas e Integrais
por Raphaela_sf » Qui Abr 05, 2012 19:26
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![\frac{\sqrt[]{\sqrt[4]{8}+\sqrt[]{\sqrt[]{2}-1}}-\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}-1}}}{\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}+1}}}](/latexrender/pictures/981987c7bcdf9f8f498ca4605785636a.png "\frac{\sqrt[]{\sqrt[4]{8}+\sqrt[]{\sqrt[]{2}-1}}-\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}-1}}}{\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}+1}}}")
![\sqrt[]{2}](/latexrender/pictures/f21662d1cabab6e8b273a4b6f1cd663a.png "\sqrt[]{2}")
(dica : igualar a expressão a 
e elevar ao quadrado os dois lados)