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Mathematics 10 Online
OpenStudy (anonymous):

sqrt(x+1)/sqrt(9x+1)

OpenStudy (anonymous):

tendendo ao infinito positivo

geerky42 (geerky42):

Do you know English?

OpenStudy (solomonzelman):

\(\Large\color{black}{ \bf \frac{ \sqrt{x+1} }{\sqrt{9x+1}} }\) mutliply top and bottom times √(9x+1)

OpenStudy (anonymous):

\[\begin{align*}\lim_{x\to\infty}\frac{\sqrt{x+1}}{\sqrt{9x+1}}\cdot\frac{\sqrt{9x+1}}{\sqrt{9x+1}}&=\lim_{x\to\infty}\frac{\sqrt{9x^2+10x+1}}{9x+1}\\\\\\ &=\lim_{x\to\infty}\frac{\sqrt{x^2}\sqrt{9+\dfrac{10}{x}+\dfrac{1}{x^2}}}{9x+1}\\\\\\ &=\lim_{x\to\infty}\frac{|x|\sqrt{9+\dfrac{10}{x}+\dfrac{1}{x^2}}}{9x+1}\\\\\\ &=\lim_{x\to\infty}\frac{x\sqrt{9+\dfrac{10}{x}+\dfrac{1}{x^2}}}{9x+1}\\\\\\ &=\lim_{x\to\infty}\frac{\sqrt{9+\dfrac{10}{x}+\dfrac{1}{x^2}}}{9+\dfrac{1}{x}} \end{align*}\] Temos \(|x|=x\) porque \(x\to\infty\), então sabemos que \(x>0\).

geerky42 (geerky42):

Hooray to bilingual user!

OpenStudy (anonymous):

limite x-> ∞ x/sqrt x+1

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