Math Problem Statement

Find the limit as x approaches 0 of (cos(3x) - 1) / (5x + tan(x))

Solution

To solve this limit problem, we need to find:

limx0cos(3x)15x+tan(x)\lim_{{x \to 0}} \frac{\cos(3x) - 1}{5x + \tan(x)}

Step-by-Step Solution

  1. Evaluate the form: As x0x \to 0, we can see that cos(3x)1\cos(3x) \to 1, tan(x)0\tan(x) \to 0, and 5x05x \to 0. So, the expression takes the indeterminate form 00\frac{0}{0}.

  2. Apply L'Hôpital's Rule: Since the limit is an indeterminate form of type 00\frac{0}{0}, we can apply L'Hôpital's Rule, which states that: limx0f(x)g(x)=limx0f(x)g(x)\lim_{{x \to 0}} \frac{f(x)}{g(x)} = \lim_{{x \to 0}} \frac{f'(x)}{g'(x)} if the limit exists. Here:

    • f(x)=cos(3x)1f(x) = \cos(3x) - 1
    • g(x)=5x+tan(x)g(x) = 5x + \tan(x)
  3. Differentiate the Numerator f(x)f(x): f(x)=ddx(cos(3x)1)=3sin(3x)f'(x) = \frac{d}{dx}(\cos(3x) - 1) = -3 \sin(3x)

  4. Differentiate the Denominator g(x)g(x): g(x)=ddx(5x+tan(x))=5+sec2(x)g'(x) = \frac{d}{dx}(5x + \tan(x)) = 5 + \sec^2(x)

  5. Rewrite the Limit: Substitute f(x)f'(x) and g(x)g'(x) back into the limit: limx0cos(3x)15x+tan(x)=limx03sin(3x)5+sec2(x)\lim_{{x \to 0}} \frac{\cos(3x) - 1}{5x + \tan(x)} = \lim_{{x \to 0}} \frac{-3 \sin(3x)}{5 + \sec^2(x)}

  6. Evaluate the New Limit: As x0x \to 0:

    • sin(3x)3x\sin(3x) \approx 3x, so 3sin(3x)9x-3 \sin(3x) \approx -9x.
    • sec2(x)1\sec^2(x) \to 1, so 5+sec2(x)65 + \sec^2(x) \to 6.

    This gives: limx09x6=96x=0\lim_{{x \to 0}} \frac{-9x}{6} = \frac{-9}{6} x = 0

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Math Problem Analysis

Mathematical Concepts

Calculus
Limits
Trigonometric Functions
Indeterminate Forms

Formulas

L'Hôpital's Rule
Trigonometric limit sin(x)/x as x approaches 0

Theorems

L'Hôpital's Rule

Suitable Grade Level

Grades 11-12