[Limites]

por **Subnik** » Sex Abr 03, 2015 19:43

Calcule o limite:
$\lim_{x\rightarrow+/-\infty}\sqrt[]{x^2-x.\Pi}-\sqrt[]{x^2-1}$

Resposta: $+/- \frac{\Pi}{2}$

por **adauto martins** » Sáb Abr 04, 2015 12:14

$L=\lim_{x\rightarrow \infty}{x}^{2}(\sqrt[]{1-\pi/x}-\sqrt[]{1-1/{x}^{2}})$ = $\lim_{x\rightarrow \infty}{x}^{2}(\sqrt[]{1-\pi/x}-\sqrt[]{1-1/{x}^{2}}).(\sqrt[]{1-\pi/x}+\sqrt[]{1-1/{x}^{2}}/(\sqrt[]{1+\pi/x}+\sqrt[]{1-1/{x}^{2}})$ = $\lim_{x\rightarrow \infty}{x}^{2}(1-\pi/x-1-1/{x}^{2})/\sqrt[]{1-\pi/x}+\sqrt[]{1-1/{x}^{2}})=\lim_{x\rightarrow \infty}-(\pi.x+1)/(\sqrt[]{1-\pi/x}+\sqrt[]{1-1/{x}^{2}})=$ = $\lim_{x\rightarrow \infty}-( \pi x + 1)/\sqrt[]{1-\pi/x}+\sqrt[]{1-1/{x}^{2}})$ = $\lim_{x\rightarrow \infty}-x(\pi+1/x)/(\sqrt[]{1-\pi/x}+\sqrt[]{1-1/{x}^{2}})$ = $\lim_{x\rightarrow \infty}-(\pi+1/{x}^{2})/(\sqrt[]{1/{x}^{2}-\pi/{x}^{3}}+\sqrt[]{1-1/{x}^{4}}=-\pi/2$

por **DanielFerreira** » Sáb Abr 04, 2015 12:23

Olá Subnik,
seja bem-vindo!

$\\ \lim_{x \to \infty} \sqrt{x^2 - x \cdot \pi} - \sqrt{x^2 - 1} = \\\\\\ \lim_{x \to \infty} \sqrt{x^2 - x \cdot \pi} - \sqrt{x^2 - 1} \times \frac{\sqrt{x^2 - x \cdot \pi} + \sqrt{x^2 - 1}}{\sqrt{x^2 - x \cdot \pi} + \sqrt{x^2 - 1}} = \\\\\\ \lim_{x \to \infty}\frac{x^2 - x \cdot \pi - (x^2 - 1)}{\sqrt{x^2 - x \cdot \pi} + \sqrt{x^2 - 1}} = \\\\\\ \lim_{x \to \infty}\frac{\cancel{x^2} - x \cdot \pi - \cancel{x^2} + 1}{\sqrt{x^2 \left ( 1 - \frac{\pi}{x} \right )} + \sqrt{x^2 \left ( 1 - \frac{1}{x^2} \right )}} = \\\\\\ \lim_{x \to \infty}\frac{- x \cdot \pi + 1}{x \cdot \sqrt{\left ( 1 - \frac{\pi}{x} \right )} + x \cdot \sqrt{\left ( 1 - \frac{1}{x^2} \right )}} =$

$\\ \lim_{x \to \infty}\frac{\cancel{x} \left ( - \pi + \frac{1}{x} \right )}{\cancel{x} \left ( \sqrt{1 - \frac{\pi}{x}} + \sqrt{1 - \frac{1}{x^2}} \right )} = \\\\\\ \lim_{x \to \infty}\frac{- \pi + \frac{1}{x}}{\sqrt{1 - \frac{\pi}{x}} + \sqrt{1 - \frac{1}{x^2}}} = \\\\\\ \frac{- \pi + 0}{\sqrt{1 - 0} + \sqrt{1 - 0}} = \\\\\\ \frac{- \pi}{1 + 1} = \\\\\\ \boxed{- \frac{\pi}{2}}$

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