To evaluate the limit of $\frac{|\sin x|}{x}$ as $x \to 0$, let's break it down step-by-step:

Step 1: Recognize the problem with direct substitution

Directly substituting $x = 0$ into $\frac{|\sin x|}{x}$ leads to an indeterminate form, as the numerator $|\sin x|$ approaches 0 and the denominator $x$ approaches 0. Thus, we need to analyze the behavior as $x$ approaches 0 from both sides.

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Step 2: Behavior of $|\sin x|$

The absolute value of $\sin x$ satisfies:

\sin x & \text{if } x > 0, \\ -\sin x & \text{if } x < 0. \end{cases}$$ For both cases, the behavior of $$\sin x$$ is bounded between $$-1$$ and $$1$$, and $$|\sin x|$$ approaches 0 as $$x \to 0$$. --- ### Step 3: Analyzing the left-hand and right-hand limits #### Case 1: $$x > 0$$ If $$x > 0$$, the limit becomes: $$\lim_{x \to 0^+} \frac{\sin x}{x}.$$ Using the standard limit $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$, we find: $$\lim_{x \to 0^+} \frac{|\sin x|}{x} = \lim_{x \to 0^+} \frac{\sin x}{x} = 1.$$ #### Case 2: $$x < 0$$ If $$x < 0$$, the limit becomes: $$\lim_{x \to 0^-} \frac{-\sin x}{x}.$$ This simplifies to: $$\lim_{x \to 0^-} \frac{\sin x}{x} = 1.$$ --- ### Step 4: Conclude Since the left-hand and right-hand limits are equal: $$\lim_{x \to 0} \frac{|\sin x|}{x} = 1.$$ --- ### Final Answer: $$\boxed{1}$$ Would you like me to elaborate on any step or explore related concepts? ### Related Questions: 1. How would the limit change if $$|\sin x|$$ were replaced by $$|x \sin x|$$? 2. What is the limit of $$\frac{\sin^2 x}{x}$$ as $$x \to 0$$? 3. How does the behavior of $$|\sin x|/x$$ compare as $$x \to \infty$$? 4. Can this result be generalized to other trigonometric functions? 5. What happens if the denominator $$x$$ is replaced by $$x^2$$? ### Tip: For any trigonometric function, remember that limits involving $$\sin x$$ and small-angle approximations often leverage the fundamental limit $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$.