## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x\to 2}\sqrt{\frac{2x^2+1}{3x-2}}=\frac{3}{2}$
$\lim\limits_{x\to 2}\sqrt{\frac{2x^2+1}{3x-2}}$ $=\sqrt{\lim\limits_{x\to 2}\frac{2x^2+1}{3x-2}}$ (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, $$\lim\limits_{x\to 2}\sqrt{\frac{2x^2+1}{3x-2}}=\frac{3}{2}$$