Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each limit. $$ \lim _{t \rightarrow 0} \frac{\cos ^{2} t}{1+\sin t} $$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Substitute the Limit Value

First, substitute the limit value, which is 0, into the expression: \( \frac{\cos^2(0)}{1+\sin(0)} \). Since \( \cos(0) = 1 \) and \( \sin(0) = 0 \), this results in \( \frac{1^2}{1+0} = 1 \).

Step 2: Verify Using Properties of Limits

To ensure the previous step is correct, we apply the properties of limits. The limit of a product is the product of the limits, and the limit of a quotient is the quotient of the limits (as long as the limit in the denominator isn't zero). Here both numerator and denominator are continuous functions at \( t = 0 \).

Step 3: Evaluate the Continuity

We confirm the continuity of \( \cos^2(t) \) and \( 1 + \sin(t) \) at \( t =0 \). Since trigonometric functions are continuous everywhere, we can directly evaluate the expressions at \( t = 0 \). Thus, \( \lim_{t \to 0} \cos^2(t) = 1 \), and \( \lim_{t \to 0} (1+\sin(t)) = 1 \).

Step 4: Combine the Results

Applying the results from the continuity properties, the expression evaluates to \( \frac{1}{1} = 1 \). Therefore, the limit is \( 1 \).

Key Concepts

Trigonometric Limits

Trigonometric limits are particularly important when dealing with calculus problems that involve trigonometric functions like sine and cosine. These functions have specific behaviors as inputs approach certain values, such as zero or infinity. An essential aspect to understanding these limits is knowing the basic properties of trigonometric functions:

• The limit of sine as it approaches zero is zero, i.e., \( \lim _{t \rightarrow 0} \sin t = 0 \).
• The limit of cosine as it approaches zero is one, i.e., \( \lim _{t \rightarrow 0} \cos t = 1 \).
• These functions are periodic and continuous over their entire domain.

In the original solution, the limit \( \lim _{t \rightarrow 0} \frac{\cos ^{2} t}{1+\sin t} \) utilized these foundational limits to substitute the expression components directly, leading to an easy calculation.

Continuity

The concept of continuity is pivotal in calculus and refers to whether a function is smooth without any interruptions or breaks at a point. For a function to be continuous at a point, three conditions must be satisfied:

• The function is defined at the point.
• The limit of the function exists as it approaches that point.
• The limit of the function equals the function's value at that point.

Trigonometric functions like \( \cos(t) \) and \( \sin(t) \) are continuous for all real numbers, enabling direct evaluation at any point, like \( t = 0 \). In the problem's solution, continuity confirms that evaluating the limit through substitution is valid because there are no sudden jumps or undefined points at \( t = 0 \). Therefore, using the properties of continuity helps ensure that the calculations are reliable and accurate.

Limit Evaluation Techniques

Calculus provides various techniques to evaluate limits, especially when expressions become complex or involve indeterminate forms. In this exercise, the limit evaluation technique of direct substitution was suitable due to the continuity of the involved trigonometric functions. However, there are several other methods available:

• Factoring: Used to cancel out common terms in cases where substitution leads to a \( \frac{0}{0} \) indeterminate form.
• Rationalization: Utilizes multiplying by a conjugate to simplify roots and tackle indeterminate forms as well.
• L'Hôpital's Rule: Applies to indeterminate forms by differentiating the numerator and denominator separately.
• Limit Laws: Utilize properties like the sum, product, and quotient of limits.

By using the simplicity of direct substitution in this problem, we confirmed accurate limit values without needing advanced methods. Understanding when and how to use each of these techniques is essential for solving a variety of limit problems in calculus.