Chapter 13 - A Preview of Calculus: The Limit, Derivative, and Integral of a Function - Section 13.2 Algebra Techniques for Finding Limits - 13.2 Assess Your Understanding - Page 903: 24

$1$

When $f(x) \geq 0$ then, we have $\lim\limits_{x \to a} \sqrt {k(x)}=\sqrt {\lim\limits_{x \to a} k(a)}$ where $a$ is a constant. Thus, we have: $\lim\limits_{x \to 0} \sqrt {1-2x}=\sqrt {\lim\limits_{x \to 0} (1-2x)} \\=\sqrt {1-(2)(0)} \\=\sqrt 1 \\=1$