Problem 44 - Thomas Calculus

Question

Limits with trigonometric functions Find the limits in Exercises \(43-50 .\) $$\lim _{x \rightarrow \pi / 4} \sin ^{2} x$$

Step-by-Step Solution

Answer

The limit is \( \frac{1}{2} \).

Step 1: Understand the Limit Expression

The given expression is \( \lim_{x \rightarrow \pi/4} \sin^2 x \), which means we want to find the value that \( \sin^2 x \) approaches as \( x \) gets closer to \( \pi/4 \).

Step 2: Substitute the Limit Value

Since the expression \( \sin^2 x \) is continuous, we can substitute \( x = \pi/4 \) directly into the expression to calculate the limit. This is based on the property that limits of continuous functions can be evaluated by direct substitution.

Step 3: Evaluate the Sine Function

Substitute \( x = \pi/4 \) into \( \sin x \). We have \( \sin(\pi/4) = \frac{\sqrt{2}}{2} \). The sine of \( \pi/4 \) is a well-known trigonometric value that equals \( \frac{\sqrt{2}}{2} \).

Step 4: Square the Sine Value

Now, square the result from the previous step: \( \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} = \frac{1}{2} \).

Step 5: Conclusion about the Limit

The limit of \( \sin^2 x \) as \( x \) approaches \( \pi/4 \) is simply the squared value of \( \sin(\pi/4) \), because \( \sin^2 x \) is continuous.

Key Concepts

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, play a vital role in mathematics, particularly in calculus. These functions are periodic, meaning they repeat their values in regular intervals. This characteristic can help in understanding and calculating limits when dealing with trigonometric expressions.
For example, the sine function, denoted as \( \sin(x) \), is fundamental in analyzing oscillating phenomena. A trigonometric limit like \( \lim_{x \rightarrow \pi/4} \sin^2 x \) involves evaluating how the function behaves as \( x \) approaches a specific angle measured in radians—in this case, \( \pi/4 \).
Key trigonometric values are critical for solving limits. Memorizing values like \( \sin(\pi/4) = \frac{\sqrt{2}}{2} \) helps in quickly computing results when limits involve common angles. Such functions are often encountered in both theoretical problems and real-world applications.

Continuous Functions

Continuous functions are at the heart of calculus and mathematical analysis. A function is continuous at a point if it behaves predictably without any jumps, breaks, or holes near that point.
This means, for any very small change in input, there's an equally small change in output. For trigonometric functions like \( \sin x \), continuity is one of their essential properties.
When a function is continuous at a point, we can find limits using direct substitution. This makes evaluating limits straightforward. For instance, since \( \sin x \) is continuous at \( x = \pi/4 \), we can substitute directly into the expression \( \sin^2 x \) to find the limit.
A continuous function allows for a predictable behavior that is crucial for smooth curve analysis and function approximation. Understanding the continuity of trigonometric functions simplifies both theoretical calculations and practical applications.

Direct Substitution

Direct substitution is a method used to evaluate limits of continuous functions efficiently. It involves plugging the value that \( x \) approaches directly into the function. This technique only works seamlessly with functions that are continuous at the concerned point.
In the problem given, since \( \sin^2 x \) is continuous at \( x = \pi/4 \), we simply substitute \( \pi/4 \) in for \( x \) to calculate the limit: \( \sin^2(\pi/4) = \left(\frac{\sqrt{2}}{2}\right)^2 \).
Remembering that direct substitution is valid because the function lacks any undefined points or discontinuities around the point allows us to complete the procedure quickly and efficiently.