To solve the limit \[ \lim_{x \to 0} \frac{\sqrt{1 - \cos(2x)}}{x} \] we can follow these steps: Step 1: Use the identity for \( \cos(2x) \) We know that: \[ \cos(2x) = 1 - 2\sin^2(x) \] Therefore, we can rewrite \( 1 - \cos(2x) \) as: \[ 1 - \cos(2x) = 1 - (1 - 2\sin^2(x)) = 2\sin^2(x) \] Step 2: Substitute into the limit Now, we can substitute this back into our limit: \[ \lim_{x \to 0} \frac{\sqrt{1 - \cos(2x)}}{x} = \lim_{x \to 0} \frac{\sqrt{2\sin^2(x)}}{x} \] Step 3: Simplify the expression The square root can be simplified as: \[ \sqrt{2\sin^2(x)} = \sqrt{2} \cdot |\sin(x)| \] $$ \begin{aligned} &= \lim_{x \to 0^+} \frac{\sqrt{2} |\sin x|}{x} = \sqrt{2} \cdot \lim_{x \to 0^+} \frac{|\sin x|}{x} = \sqrt{2} \cdot \lim_{x \to 0^+} \frac{\sin x}{x} = \sqrt{2} \cdot 1 = \boxed{\sqrt{2}} \\[10pt] &\text{and} \\[10pt] &= \lim_{x \to 0^-} \frac{\sqrt{2} |\sin x|}{x} = \sqrt{2} \cdot \lim_{x \to 0^-} \frac{|\sin x|}{x} = \sqrt{2} \cdot \lim_{x \to 0^-} \frac{-\sin x}{x} = \sqrt{2} \cdot (-1) = \boxed{-\sqrt{2}} \end{aligned} $$ Hence, the limit does not exist as left-hand and right-hand limits are not equal.