limit as x approaches 0 of (tan(2x))/(6xsec(3x))

We are tasked with evaluating the limit:

\[
\lim_{x \to 0} \frac{\tan(2x)}{6x \sec(3x)}
\]

### Step 1: Break down the expression
We can rewrite the expression as:

\[
\frac{\tan(2x)}{6x \sec(3x)} = \frac{\frac{\sin(2x)}{\cos(2x)}}{6x \cdot \frac{1}{\cos(3x)}} = \frac{\sin(2x)}{6x \cos(2x) \cos(3x)}
\]

Now the limit becomes:

\[
\lim_{x \to 0} \frac{\sin(2x)}{6x \cos(2x) \cos(3x)}
\]

### Step 2: Apply standard limits
We use the standard limit \( \lim_{x \to 0} \frac{\sin(kx)}{kx} = 1 \), where \( k \) is a constant.

Here, we have \( \frac{\sin(2x)}{2x} \), so we multiply and divide by 2:

\[
\lim_{x \to 0} \frac{\sin(2x)}{6x \cos(2x) \cos(3x)} = \lim_{x \to 0} \frac{2}{6} \cdot \frac{\sin(2x)}{2x} \cdot \frac{1}{\cos(2x) \cos(3x)}
\]

\[
= \frac{2}{6} \cdot 1 \cdot \frac{1}{\cos(0) \cos(0)} = \frac{2}{6} \cdot 1 = \frac{1}{3}
\]

### Final Answer:
\[
\lim_{x \to 0} \frac{\tan(2x)}{6x \sec(3x)} = \frac{1}{3}
\]

Would you like further clarification or more details?

Here are some related questions:
1. How does the small-angle approximation for trigonometric functions help in solving limits?
2. How does the limit change if the trigonometric argument is a different constant?
3. What are some common trigonometric limits used in calculus?
4. Can L'Hôpital's Rule be applied to this limit? If so, how would it work?
5. How does this problem differ if higher-order terms are considered?

**Tip:** When working with trigonometric limits, always consider using small-angle approximations like \( \sin(x) \approx x \) and \( \tan(x) \approx x \) as \( x \to 0 \).

Explore how to evaluate the limit of (tan(2x))/(6xsec(3x)) as x approaches 0 using trigonometric identities and small-angle approximations. Learn the key concepts and formulas for solving limits in calculus. Suitable for Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation