Answer
$\lim\limits_{x\to 2}\sqrt{\frac{2x^2+1}{3x-2}}=\frac{3}{2}$
Work Step by Step
$\lim\limits_{x\to 2}\sqrt{\frac{2x^2+1}{3x-2}}$
$=\sqrt{\lim\limits_{x\to 2}\frac{2x^2+1}{3x-2}}$ (root law)
$=\sqrt A$ (1)
We need to check if $A\geq0$ so that $\sqrt A$ is defined.
$A=\lim\limits_{x\to 2}\frac{2x^2+1}{3x-2}$
$A=\frac{\lim\limits_{x\to 2}(2x^2+1)}{\lim\limits_{x\to 2}(3x-2)}$ (quotient law) (let's assume that $\lim\limits_{x\to 2}(3x-2)\ne0$)
$A=\frac{\lim\limits_{x\to 2}2x^2+\lim\limits_{x\to 2}1}{\lim\limits_{x\to 2}3x-\lim\limits_{x\to 2}2}$ (sum and difference law)
$A=\frac{2\lim\limits_{x\to 2}x^2+\lim\limits_{x\to 2}1}{3\lim\limits_{x\to 2}x-\lim\limits_{x\to 2}2}$ (constant multiple law)
$A=\frac{2\times2^2+1}{3\times2-2}$
$A=\frac{9}{4}\gt0$
Therefore, $\sqrt A$ is defined.
Continue with (1), we have
$\sqrt A=\sqrt{\frac{9}{4}}=\frac{3}{2}$
In conclusion, $$\lim\limits_{x\to 2}\sqrt{\frac{2x^2+1}{3x-2}}=\frac{3}{2}$$
