sqrt(x+1)/sqrt(9x+1)
tendendo ao infinito positivo
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\(\Large\color{black}{ \bf \frac{ \sqrt{x+1} }{\sqrt{9x+1}} }\) mutliply top and bottom times √(9x+1)
\[\begin{align*}\lim_{x\to\infty}\frac{\sqrt{x+1}}{\sqrt{9x+1}}\cdot\frac{\sqrt{9x+1}}{\sqrt{9x+1}}&=\lim_{x\to\infty}\frac{\sqrt{9x^2+10x+1}}{9x+1}\\\\\\ &=\lim_{x\to\infty}\frac{\sqrt{x^2}\sqrt{9+\dfrac{10}{x}+\dfrac{1}{x^2}}}{9x+1}\\\\\\ &=\lim_{x\to\infty}\frac{|x|\sqrt{9+\dfrac{10}{x}+\dfrac{1}{x^2}}}{9x+1}\\\\\\ &=\lim_{x\to\infty}\frac{x\sqrt{9+\dfrac{10}{x}+\dfrac{1}{x^2}}}{9x+1}\\\\\\ &=\lim_{x\to\infty}\frac{\sqrt{9+\dfrac{10}{x}+\dfrac{1}{x^2}}}{9+\dfrac{1}{x}} \end{align*}\] Temos \(|x|=x\) porque \(x\to\infty\), então sabemos que \(x>0\).
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limite x-> ∞ x/sqrt x+1
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