https://file.aiquickdraw.com/mathbot/file/1727410747033kpyrw23f.jpg

Kerjakan soal berikut menggunakan rumus limit trigonometri

The limit problem presented in the image is:

\[
\lim_{{x \to 0}} \frac{\sin(8x)}{\tan(9x)}
\]

We can solve this using trigonometric limit rules. One useful rule is:

\[
\lim_{{x \to 0}} \frac{\sin(kx)}{kx} = 1 \quad \text{and} \quad \lim_{{x \to 0}} \frac{\tan(kx)}{kx} = 1
\]

We can rewrite the given limit by factoring constants from both the sine and tangent terms:

\[
\lim_{{x \to 0}} \frac{\sin(8x)}{\tan(9x)} = \lim_{{x \to 0}} \frac{\frac{\sin(8x)}{x}}{\frac{\tan(9x)}{x}}
\]

Using the trigonometric limits:

\[
\lim_{{x \to 0}} \frac{\sin(8x)}{x} = 8 \quad \text{and} \quad \lim_{{x \to 0}} \frac{\tan(9x)}{x} = 9
\]

Thus, the limit becomes:

\[
\frac{8}{9}
\]

So the solution is:

\[
\lim_{{x \to 0}} \frac{\sin(8x)}{\tan(9x)} = \frac{8}{9}
\]

Would you like further details or clarification? Here are some related questions:

1. What is the general form of the trigonometric limit rule?
2. How can this approach be applied to other trigonometric limit problems?
3. Why are small-angle approximations useful in limit calculations?
4. Can you explain why \(\lim_{{x \to 0}} \frac{\sin(kx)}{kx} = 1\)?
5. What happens if different coefficients are used for \(x\) in sine and tangent?

**Tip:** Recognizing standard limits such as \(\lim_{{x \to 0}} \frac{\sin(x)}{x} = 1\) greatly simplifies limit problems involving trigonometric functions.

Solve the limit of sin(8x)/tan(9x) as x approaches 0 using trigonometric limit rules. This problem involves the use of the standard limit theorem and applies to advanced trigonometry and calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation