Math Problem Statement

Kerjakan soal berikut menggunakan rumus limit trigonometri

Kerjakan soal berikut menggunakan rumus limit trigonometri: lim(x -> 0) sin(2x) / 3x

Solution

The problem presented is a limit involving a trigonometric function:

limx0sin(2x)3x\lim_{{x \to 0}} \frac{\sin(2x)}{3x}

To solve this, we will use the standard limit identity for small angles:

limx0sin(kx)kx=1\lim_{{x \to 0}} \frac{\sin(kx)}{kx} = 1

where kk is a constant. In our case, the numerator is sin(2x)\sin(2x), so k=2k = 2, and the denominator is 3x3x.

Here are the steps:

  1. Rewrite the limit as: limx0sin(2x)3x=13limx0sin(2x)2x×2\lim_{{x \to 0}} \frac{\sin(2x)}{3x} = \frac{1}{3} \lim_{{x \to 0}} \frac{\sin(2x)}{2x} \times 2

  2. Apply the standard limit rule: limx0sin(2x)2x=1\lim_{{x \to 0}} \frac{\sin(2x)}{2x} = 1

  3. Therefore, the limit becomes: 13×1×2=23\frac{1}{3} \times 1 \times 2 = \frac{2}{3}

The solution to the limit is:

limx0sin(2x)3x=23\lim_{{x \to 0}} \frac{\sin(2x)}{3x} = \frac{2}{3}

Let me know if you need more details or have any questions.

Here are five related questions to expand on this concept:

  1. What is the limit of sin(3x)4x\frac{\sin(3x)}{4x} as x0x \to 0?
  2. How would you evaluate limx0sin(x)5x\lim_{{x \to 0}} \frac{\sin(x)}{5x}?
  3. Can you use L'Hôpital's Rule to solve limx0sin(kx)mx\lim_{{x \to 0}} \frac{\sin(kx)}{mx}?
  4. What is the general form of the small angle approximation for trigonometric functions?
  5. How does the limit sin(x)x\frac{\sin(x)}{x} behave for larger angles?

Tip: Always check for simplifications involving trigonometric identities before applying limit theorems.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Limits
Calculus

Formulas

lim(x -> 0) (sin(kx)/kx) = 1
lim(x -> 0) sin(2x) / 3x = (1/3) lim(x -> 0) sin(2x)/2x

Theorems

Trigonometric Limit Theorem
Small Angle Approximation

Suitable Grade Level

Grades 10-12