Math Problem Statement

Evaluate the limit \(\lim_{x \to \pi} \frac{|2x - 2\pi|}{\sin(2x)}\).

Solution

The given problem is:

lim⁡x→π∣2x−2π∣sin⁡(2x).\lim_{x \to \pi} \frac{|2x - 2\pi|}{\sin(2x)}.

Let us solve it step by step.

Step 1: Simplify the numerator and denominator

The numerator is ∣2x−2π∣|2x - 2\pi|, which represents the absolute value of 2x−2π2x - 2\pi. The denominator is sin⁡(2x)\sin(2x).

Step 2: Behavior of numerator and denominator as x→πx \to \pi

  • As x→πx \to \pi, the term 2x−2π→02x - 2\pi \to 0. So the numerator approaches 00.
  • For the denominator, sin⁡(2x)\sin(2x), when x=πx = \pi, 2x=2π2x = 2\pi, and sin⁡(2π)=0\sin(2\pi) = 0. Therefore, the denominator also approaches 00.

Since both the numerator and denominator approach 00, we can apply L’Hôpital’s Rule to evaluate this limit.

Step 3: Differentiate numerator and denominator

Differentiating the numerator ∣2x−2π∣|2x - 2\pi|: ddx∣2x−2π∣=sgn(2x−2π)⋅2,\frac{d}{dx} |2x - 2\pi| = \text{sgn}(2x - 2\pi) \cdot 2, where sgn(2x−2π)\text{sgn}(2x - 2\pi) is the sign function, which equals:

\begin{cases} 1, & \text{if } 2x - 2\pi > 0, \\ -1, & \text{if } 2x - 2\pi < 0. \end{cases}$$ So: $$\frac{d}{dx} |2x - 2\pi| = 2 \quad \text{(since $$2x - 2\pi > 0$$ as $$x \to \pi^+$$)}.$$ Differentiating the denominator $$\sin(2x)$$: $$\frac{d}{dx} \sin(2x) = 2\cos(2x).$$ --- ### Step 4: Apply L’Hôpital’s Rule By L’Hôpital’s Rule, the limit becomes: $$\lim_{x \to \pi} \frac{|2x - 2\pi|}{\sin(2x)} = \lim_{x \to \pi} \frac{\frac{d}{dx}|2x - 2\pi|}{\frac{d}{dx}\sin(2x)} = \lim_{x \to \pi} \frac{2}{2\cos(2x)}.$$ --- ### Step 5: Simplify the result At $$x = \pi$$, we calculate $$\cos(2x)$$: $$2x = 2\pi, \quad \cos(2\pi) = 1.$$ So the limit becomes: $$\frac{2}{2 \cdot 1} = 1.$$ --- ### Final Answer: $$\lim_{x \to \pi} \frac{|2x - 2\pi|}{\sin(2x)} = 1.$$ --- Would you like a detailed explanation of any step or have further questions? --- ### Related Questions: 1. What are the conditions for applying L’Hôpital’s Rule? 2. How do you differentiate absolute value functions in general? 3. What happens if the limit approaches infinity instead of zero? 4. Can this problem be solved using Taylor series expansion? 5. Why does $$\sin(2\pi) = 0$$, and how does it affect the denominator? --- ### Tip: When dealing with absolute values in limits, always check whether the expression inside the absolute value is positive or negative in the neighborhood of the limit point. This helps in simplifying the problem correctly.

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Math Problem Analysis

Mathematical Concepts

Limits
Absolute Value
Trigonometric Functions
Differentiation

Formulas

L’Hôpital’s Rule: \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \text{ if } f(c) = g(c) = 0 \)
Derivative of \( \sin(2x) \): \( \frac{d}{dx} \sin(2x) = 2 \cos(2x) \)
Derivative of \( |u| \): \( \frac{d}{dx}|u| = \text{sgn}(u) \cdot u' \)

Theorems

L’Hôpital’s Rule

Suitable Grade Level

Grades 11-12

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