Problem 9 - Calculus Early Transcendentals

Question

Use Theorem 3.11 to evaluate the following limits. $$\lim _{x \rightarrow 0} \frac{\sin 7 x}{\sin 3 x}$$

Step-by-Step Solution

Answer

#Answer# The limit of the given function as x approaches 0 is $\frac{7}{3}$.

Step 1: Rewrite the function

Rewrite the limit function as: $$\lim _{x \rightarrow 0} \frac{\sin 7x}{\sin 3x} = \lim _{x \rightarrow 0} \frac{\sin 7x}{x} \cdot \frac{x}{\sin 3x}$$

Step 2: Apply Theorem 3.11

Apply Theorem 3.11 to both limits separately: $$\lim _{x \rightarrow 0} \frac{\sin 7x}{x} = 7$$ and $$\lim _{x \rightarrow 0} \frac{x}{\sin 3x} = \frac{1}{3}$$ (using the reciprocal of the theorem for the second limit)

Step 3: Combine the results

Multiply the results of Step 2 together to find the final limit: $$7 \cdot \frac{1}{3} = \frac{7}{3}$$ The limit of the given function as x approaches 0 is: $$\lim _{x \rightarrow 0} \frac{\sin 7 x}{\sin 3 x} = \frac{7}{3}$$

Key Concepts

Trigonometric Limits

Trigonometric limits focus on determining the behavior of trigonometric functions, like sine or cosine, as they approach specific values. These types of limits are important because they frequently appear in calculus, especially when dealing with periodic functions or when analyzing oscillatory motion.

When approaching limits involving trigonometric functions, we often consider the limit of \(\sin x / x\) as \(x\) approaches zero. A fundamental result is:

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

These results help simplify and analyze complex trigonometric expressions.

In our exercise, we rewrote the original function to utilize a similar logic. By breaking down \( \frac{\sin 7x}{\sin 3x} \) as a product of \( \frac{\sin 7x}{x} \cdot \frac{x}{\sin 3x} \), we can tackle each limit separately, which makes the computations easier. This approach uses the property of multiplying two known trigonometric limits, achieving a more complex result in a straightforward manner.

Theorem 3.11

Theorem 3.11 is a pivotal result in calculus, particularly for trigonometric limits. The theorem asserts:

• \(\lim_{x \to 0} \frac{\sin ax}{ax} = 1\)

where \(a\) is a constant.

This theorem is essential when analyzing trigonometric expressions where a scalar multiple inside the trigonometric function influences the behavior of the limit.

In our step-by-step solution, Theorem 3.11 allowed us to evaluate the limits \(\lim_{x \to 0} \frac{\sin 7x}{x} = 7\) and \(\lim_{x \to 0} x / \sin 3x = \frac{1}{3}\) separately.

• By recognizing the form of the theorem in the expression, it facilitated simplifying the limit to a manageable constant value.

Through this approach, the theorem reduces the complexity of handling trigonometric expressions near zero, transforming them into simple numerical evaluations.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful technique in calculus for evaluating limits that result in indeterminate forms such as \(0/0\) or \(\infty/\infty\).

Though not applied directly in this step-by-step solution, it's frequently employed with trigonometric limits, especially when they don't align neatly with known theorems like Theorem 3.11.

The rule states:

• If \(\lim_{x \to c} f(x) / g(x)\) yields an indeterminate form, then \(\lim_{x \to c} f(x) / g(x) = \lim_{x \to c} f'(x) / g'(x)\), provided this limit exists.

L'Hôpital's Rule is particularly useful when the limit involves differentiable functions, and it may recast a stubborn limit into one that is easily solvable.

In contexts where Theorem 3.11 cannot be directly applied, L'Hôpital’s Rule can be a viable alternative for solving limits involving trigonometric functions, reinforcing its importance in calculus as a flexible tool for analyzing limits.