## Thomas' Calculus 13th Edition

$0$, discontinuous.
Step 1. Simplify the rational function under the square root: $\frac{cos^2x-cos(x)}{x}=\frac{cos(x)(cos(x)-1)}{x}=\frac{cos(x)(-2sin^2(x/2))}{x}=-cos(x)sin(x/2)\frac{sin(x/2)}{x/2}$ Step 2. Since $\lim_{x\to0}\frac{sin(x/2)}{x/2}=1$, $\lim_{x\to0}sin(x/2)=0$, and $\lim_{x\to0}\cos(x)=1$, we have $\lim_{x\to0}\frac{cos^2x-cos(x)}{x}=0$, Step 3. We can reach a result that $\lim_{x\to0}sin \sqrt {|\frac{cos^2x-cos(x)}{x}|}=0$, Step 4. The function is not continuous at $x=0$ as it is not defined at this point (dividing by zero). Please note: we need to put absolute signs under the square root to avoid negative values.