Problem 39 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \lim _{x \rightarrow 0} \frac{\sin x / 2}{x} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Problem

We are tasked with finding the limit \( \lim _{x \rightarrow 0} \frac{\sin(x/2)}{x} \). This is a problem of evaluating a limit involving a trigonometric function.

Step 2: Recognize Special Limits

Recall the important limit \( \lim_{u \to 0} \frac{\sin u}{u} = 1 \). This is a standard limit used for similar problems involving trigonometric functions.

Step 3: Substitute and Simplify

In order to use the special limit, rewrite the expression in the limit in a form similar to \( \frac{\sin u}{u} \). Let's substitute \( u = \frac{x}{2} \). Then, \( x = 2u \).

Step 4: Express Limit in Terms of New Variable

Using the substitution \( u = \frac{x}{2} \), transform the limit: \[\lim_{x \to 0} \frac{\sin(x/2)}{x} = \lim_{x \to 0} \frac{\sin u}{2u} = \frac{1}{2} \lim_{u \to 0} \frac{\sin u}{u}\]The expression simplifies to \( \frac{1}{2} \)

Step 5: Apply Limit Result

Now apply the known limit result \( \lim_{u \to 0} \frac{\sin u}{u} = 1 \) to the expression from the previous step. It becomes:\[\frac{1}{2} \cdot 1 = \frac{1}{2}\]

Step 6: Conclude Solution

After evaluating the limit using substitution and standard limits, we find that the limit \(\lim _{x \rightarrow 0} \frac{\sin(x/2)}{x} = \frac{1}{2}.\)

Key Concepts

Trigonometric Limits

Trigonometric limits are a fundamental concept in calculus, especially when dealing with problems involving trigonometric functions.
Understanding these limits can significantly simplify the evaluation of complex expressions.
One of the most important trigonometric limits that you will frequently encounter is:

• The limit: \(\lim_{x \to 0} \frac{\sin x}{x} = 1\).

This limit is essential because it provides a powerful tool to tackle limit problems where sine is involved.
Its applications extend beyond just sine, as you'll often utilize it in conjunction with other functions through clever substitutions or modifications.
When evaluating trigonometric limits, it's crucial to remember that these often involve angles approaching zero, which makes them particularly suitable for approximation techniques such as substitution.

Substitution Method

The substitution method is a vital technique for solving limit problems, particularly when the direct approach isn't immediately clear.
This method involves changing variables to simplify the expression, making it easier to evaluate the limit.
In our exercise, we used the substitution \(u = \frac{x}{2}\) to transform the original expression into a format that allows the use of standard limits.

• When substituting, always consider what will make the expression easier, such as aligning it with known limits or standard forms.
• Substitution is especially useful when you have composite functions or when expressions are tangled or nested.

This technique is not only practical but also highlights the flexibility and creativity needed in solving calculus problems, as you may often need to try different substitutions to see which yields the best results.

Standard Limits

Standard limits are pre-established limits that are often used as tools to solve calculus problems, especially when evaluating the limits of complex functions.
They serve as benchmarks that can be directly applied or adapted using techniques like substitution.
For trigonometric functions, the most critical standard limit to memorize is:

• \(\lim_{u \to 0} \frac{\sin u}{u} = 1\), which is frequently used in various calculus problems.

Other standard limits include \(\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e\), and \(\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1\).
By knowing these limits, typical calculus problems become more manageable, as these limits provide a framework for solving problems that could otherwise require more cumbersome calculations.

Calculus Problems

Calculus problems, especially those involving limits, are foundational in understanding how changes occur within mathematical functions.
They often require several steps using various mathematical principles, such as algebraic manipulation or trigonometric identities, alongside calculus techniques.
Limit problems like the original exercise help us understand the behavior of functions as they approach certain values or points.

• Problems typically involve finding these points of continuity or discontinuity and are crucial for further topics in calculus, like derivatives and integrals.
• Understanding how to tackle various types of problems, including trigonometric, logarithmic, and polynomial limits, builds the essential skills needed for more advanced topics.

Solving these problems requires practice and often a step-by-step approach to simplify and evaluate the expressions effectively, as shown in the step-by-step solution of the original problem.