# Find the limit: lim_{xrightarrowinfty}frac{sqrt{9x+1}}{sqrt{x+1}}

Find the limit:
$\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{9x+1}}{\sqrt{x+1}}$
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Arnold Odonnell
To find the liimit,
$\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{9x+1}}{\sqrt{x+1}}$
$\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{9x+1}}{\sqrt{x+1}}=\underset{x\to \mathrm{\infty }}{lim}\sqrt{\frac{9x+1}{x+1}}$
$=\underset{x\to \mathrm{\infty }}{lim}\sqrt{\frac{x\left(9+\frac{1}{x}\right)}{x\left(1+\frac{1}{x}\right)}}$
$=\underset{x\to \mathrm{\infty }}{lim}\sqrt{\frac{\left(9+\frac{1}{x}\right)}{\left(1+\frac{1}{x}\right)}}$

$=\sqrt{9}$
$=3$
Therefore, $\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{9x+1}}{\sqrt{x+1}}=3$
Jeffrey Jordon