Answer
$3$
Work Step by Step
Here, we have $\lim\limits_{x \to \infty} \dfrac{\sqrt {9x+1}}{\sqrt {x+1}}=\sqrt{\lim\limits_{x \to \infty} [\dfrac{9x+1}{x+1}]}$
or, $ \sqrt{\lim\limits_{x \to \infty} \dfrac{9x+1}{x+1}(\dfrac{\frac{1}{x}}{\frac{1}{x}})}=\sqrt{\lim\limits_{x \to \infty} \dfrac{9+(1/x)}{1+(1/x)}}$
and
$\sqrt {\dfrac{9+0}{1+0}}=\sqrt 9=3$
