Math Problem Statement

lim x -> 0 (3 - 3cos^2 x)/(x ^ 2) adalah...

Solution

We are tasked with finding the limit:

limx033cos2xx2\lim_{x \to 0} \frac{3 - 3\cos^2 x}{x^2}

Step 1: Factor out the constant

We can factor out the constant 33 from the numerator:

limx03(1cos2x)x2\lim_{x \to 0} \frac{3(1 - \cos^2 x)}{x^2}

Step 2: Use a trigonometric identity

Recall that 1cos2x=sin2x1 - \cos^2 x = \sin^2 x (from the Pythagorean identity). Substituting this into the expression:

limx03sin2xx2\lim_{x \to 0} \frac{3\sin^2 x}{x^2}

Step 3: Simplify the limit

We know that limx0sinxx=1\lim_{x \to 0} \frac{\sin x}{x} = 1, so we can rewrite the expression:

limx03(sinxx)2=3×12=3\lim_{x \to 0} 3 \left( \frac{\sin x}{x} \right)^2 = 3 \times 1^2 = 3

Thus, the value of the limit is:

3\boxed{3}

Would you like further clarification on any step?

Here are five related questions you can explore:

  1. What is the limit of limx01cosxx2\lim_{x \to 0} \frac{1 - \cos x}{x^2}?
  2. How can we derive limx0sinxx\lim_{x \to 0} \frac{\sin x}{x} using L'Hopital's Rule?
  3. What are some common trigonometric limits and how are they derived?
  4. How does L'Hopital's Rule apply to limits involving indeterminate forms?
  5. What is the Maclaurin series for sinx\sin x and cosx\cos x, and how can it be used in limits?

Tip: For small angles, trigonometric functions like sinx\sin x and cosx\cos x can often be approximated using Taylor or Maclaurin series expansions.

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Math Problem Analysis

Mathematical Concepts

Limits
Trigonometric Identities
L'Hopital's Rule

Formulas

lim_{x \to 0} \frac{\sin x}{x} = 1
1 - \cos^2 x = \sin^2 x

Theorems

Pythagorean Identity
Trigonometric Limits

Suitable Grade Level

Grades 11-12