Problem 29 - Calculus for Biology and Medicine

Question

In Problems \(29-48\), find the limits. $$ \lim _{x \rightarrow \pi / 3} \sin \left(\frac{x}{2}\right) $$

Step-by-Step Solution

Answer

\( \lim _{x \rightarrow \pi / 3} \sin \left(\frac{x}{2}\right) = \frac{1}{2} \).

Step 1: Identify the limit expression

The given problem is to find the limit of the function \( \sin \left( \frac{x}{2} \right) \) as \( x \to \frac{\pi}{3} \). We need to determine the limits as \( x \) approaches this value.

Step 2: Substitute the limit point into the function

Direct substitution is often the first technique to apply. Substitute \( x = \frac{\pi}{3} \) into the function to find the limit: \( \sin \left( \frac{1}{2} \cdot \frac{\pi}{3} \right) \).

Step 3: Simplify the expression inside the sine function

Simplify \( \frac{1}{2} \cdot \frac{\pi}{3} \) to get \( \frac{\pi}{6} \). So, the function becomes \( \sin \left( \frac{\pi}{6} \right) \).

Step 4: Evaluate the sine function

Recall the sine value for common angles: \( \sin \left( \frac{\pi}{6} \right) = \frac{1}{2} \). Thus, the limit is \( \frac{1}{2} \).

Key Concepts

Understanding Trigonometric Limits

Trigonometric limits involve finding the limit of trigonometric functions as the variable approaches a specific value. This is an important concept in calculus, particularly when dealing with functions like sine, cosine, or tangent. The process helps us understand the behavior of these functions as they get close to certain points.

In our practice problem, we are tasked to find the limit of \( \sin \left( \frac{x}{2} \right) \) as \( x \to \frac{\pi}{3} \). Trigonometric limits require understanding the behavior of angles measured in radians, a unit often used in calculus for more precise measurement of angles. Radians link the unit circle, a valuable tool in trigonometry, with the lengths of the arcs they represent.

To tackle these limits, you typically use known values of trigonometric functions at key angles, like \( \sin(0) \), \( \sin(\pi/6) \), \( \sin(\pi/4) \), and so on. Familiarity with these values is essential for solving these types of problems seamlessly.

Using the Direct Substitution Method

The direct substitution method is one of the simplest techniques to use when calculating limits, especially if the function is continuous at the point of interest. In our exercise, we directly substitute \( x = \frac{\pi}{3} \) into the function \( \sin \left( \frac{x}{2} \right) \).

Here's a step-by-step process of how direct substitution works in this problem:

• Identify the function and the point you need to substitute. Here, it's \( \sin \left( \frac{x}{2} \right) \) and \( x \to \frac{\pi}{3} \).
• Substitute the limit value into the function: \( \sin \left( \frac{1}{2} \cdot \frac{\pi}{3} \right) \).
• Simplify the expression inside the sine function. \( \frac{1}{2} \cdot \frac{\pi}{3} = \frac{\pi}{6} \).
• Recognize \( \sin \left( \frac{\pi}{6} \right) \) as a standard angle for which you know the sine value.

If the function is continuous at the point, direct substitution will give you the limit instantly, which makes it a quick and effective method.

Evaluating Trigonometric Functions

Once you have simplified the problem to a basic trigonometric function evaluation, the key is knowing the sine, cosine, and tangent values at important angles. For \( \sin \left( \frac{\pi}{6} \right) \), this value is \( \frac{1}{2} \). These values are derived from the unit circle, which helps us visualize and calculate trigonometric values on a radius of one.

Here's a quick refresher for some common angle values:

• \( \sin \left( \frac{\pi}{6} \right) = \frac{1}{2} \)
• \( \sin \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \)
• \( \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} \)

Besides knowing these values, understanding why they are what they are can entail breaking down the unit circle. Envision a circle with a radius of 1, which allows these calculations to be based on pure geometric relationships of the circle, offering deeper insight into why these functions take on these particular values at standard angles.