Calcule o seguinte Limite( sem usar a Regra de L'hospital):.... Pergunta de ideia deRonnyMat

andresccp $\lim_{x \to 1} \frac{\sqrt{x}+\sqrt[3]{x}+\sqrt[4]{x}-3}{x-1}\\\\ = \lim_{x \to 1} \frac{\sqrt{x}+\sqrt[3]{x}+\sqrt[4]{x}-1-1-1}{x-1}\\\\ = \lim_{x \to 1} \frac{\sqrt{x}-1}{x-1}+ \lim_{x \to 1} \frac{\sqrt[3]{x}-1}{x-1} + \lim_{x \to 1} \frac{\sqrt[4]{x}-1}{x-1}\\\\$

lembrando que:
$\bmatrix a^2-b^2=(a-b)(a+b)\\\\a^3-b^3=(a-b)(a^2+ab+b^2)\\\\a^4-b^4=(a^2)^2-(b^2)^2=(a^2-b^2)(a^2+b^2)= (a-b)(a+b)(a^2+b^2) \end$

resolvendo os limites: nos três casos vc ja tem o (a-b) então
para eliminar a raiz quadrada multiplica por (a+b)

$\lim_{x \to 1} \frac{\sqrt{x}-1}{x-1} * \frac{\sqrt{x}+1}{\sqrt{x}+1} \\\\ = \lim_{x \to 1} \frac{(\sqrt{x})^2-1^2}{(x-1)(\sqrt{x}+1)} \\\\ = \lim_{x \to 1} \frac{(x-1)}{(x-1)(\sqrt{x}+1)} = \lim_{x \to 1} \frac{1}{(\sqrt{x}+1)} = \frac{1}{2}$
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para eliminar a raiz cubica multiplica por (a^2+ab+b^2)

$\lim_{x \to 1} \frac{\sqrt[3]{x}-1}{x-1} * \frac{(\sqrt[3]{x})^2+\sqrt[3]{x}+1}{{(\sqrt[3]{x})^2+\sqrt[3]{x}+1}} \\\\ = \lim_{x \to 1} \frac{( \sqrt[3]{x} )^3-1^3}{{(x-1)((\sqrt[3]{x})^2+\sqrt[3]{x}+1)}} \\\\ = \lim_{x \to 1} \frac{(x-1)}{{(x-1)((\sqrt[3]{x})^2+\sqrt[3]{x}+1)}} = \lim_{x \to 1} \frac{1}{{((\sqrt[3]{x})^2+\sqrt[3]{x}+1)}} = \frac{1}{3}$

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para eliminar a raiz quarta multiplica por (a+b)(a^2+b^2)

$\lim_{x \to 1} \frac{\sqrt[4]{x}-1}{x-1} * \frac{(\sqrt[4]{x}+1)((\sqrt[4]{x})^2+1)}{(\sqrt[4]{x}+1)((\sqrt[4]{x})^2+1)} \\\\ \lim_{x \to 1} \frac{(\sqrt[4]{x})^4-1^4}{(x-1)(\sqrt[4]{x}+1)((\sqrt[4]{x})^2+1)} \\\\ \lim_{x \to 1} \frac{x-1}{(x-1)(\sqrt[4]{x}+1)((\sqrt[4]{x})^2+1)} = \lim_{x \to 1} \frac{1}{(\sqrt[4]{x}+1)((\sqrt[4]{x})^2+1)} = \frac{1}{4}$

$lim_{x \to 1} \frac{\sqrt{x}+\sqrt[3]{x}+\sqrt[4]{x}-3}{x-1}= \frac{1}{2} + \frac{1}{3}+ \frac{1}{4} \\\\\boxed{\boxed{lim_{x \to 1} \frac{\sqrt{x}+\sqrt[3]{x}+\sqrt[4]{x}-3}{x-1}= \frac{13}{12} }}$

2 votes Thanks 2

RonnyMat C*RALHO MANO.. VC TA LOUCO KKKKKKK VALEU. AQUELE ABRACO

BashKnocker $\lim\limits_{x\to 1}\frac{\sqrt[3]{x}+\sqrt{x}+\sqrt[4]{x}-3}{x-1}=\\\\ \lim\limits_{x\to 1}\frac{\sqrt[3]{x}+\sqrt{x}+\sqrt[4]{x}-1-1-1}{x-1}=\\\\ \lim\limits_{x\to 1}\frac{(\sqrt[3]{x}-1)+(\sqrt{x}-1)+(\sqrt[4]{x}-1)}{x-1}=\\\\
\lim\limits_{x\to 1}\left[\frac{\sqrt[3]{x}-1}{x-1}+\frac{\sqrt{x}-1}{x-1}+\frac{\sqrt[4]{x}-1}{x-1}\right]=\\\\ \lim\limits_{x\to 1}\frac{\sqrt[4]{x}-1}{x-1}+\lim\limits_{x\to 1}\frac{\sqrt[3]{x}-1}{x-1}+\lim\limits_{x\to 1}\frac{\sqrt[2]{x}-1}{x-1}=$

Como o limite tende o x para 1 podemos reescrever as expressões de cada limite na forma de logaritmos na base x. Todos os termos da soma tendem ao valor do limite, ou seja, todos os limites continuam na indeterminação 0/0.

$\lim\limits_{x\to 1}\frac{\log_x(\frac{\sqrt[4]{x}}{1})}{\log_x(\frac{x}{1})}+\lim\limits_{x\to 1}}\frac{\log_x(\frac{\sqrt[3]{x}}{1})}{\log_x(\frac{x}{1})}+\lim\limits_{x\to 1}\frac{\log_x(\frac{\sqrt[2]{x}}{1})}{\log_x(\frac{x}{1})}=\\\\
\lim\limits_{x\to 1} \frac{\frac{1}{4}\log_x(x)-\log_x(1)}{\log_x(x)-\log_x(1)} + \lim\limits_{x\to 1}\frac{\frac{1}{3}\log_x(x)-\log_x(1)}{\log_x(x)-\log_x(1)} + \lim\limits_{x\to 1}\frac{\frac{1}{2}\log_x(x)-\log_x(1)}{\log_x(x)-\log_x(1)}=\\\\ \lim\limits_{x\to 1}\frac{\frac{1}{4}\times 1 - 0}{1-0}+\lim\limits_{x\to 1}\frac{\frac{1}{3}\times 1 - 0}{1-0}+\lim\limits_{x\to 1}\frac{\frac{1}{2}\times 1 - 0}{1-0}=\\\\ \lim_{x \to 1}\frac{1}{4}+\lim_{x \to 1}\frac{1}{3}+\lim_{x \to 1}\frac{1}{2}=\\\\
\frac{1}{4}+\frac{1}{3}+\frac{1}{2}=\\\\ \boxed{\frac{13}{12}}$

2 votes Thanks 3

RonnyMat MUITO BEM JOGADO MANO !!!

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