Question

True or False? In Exercises \(115-120\) , determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. $$ \lim _{x \rightarrow \pi} \frac{\sin x}{x}=1 $$

Short answer

The statement is false as the limit \(\lim_{x \rightarrow \pi} \frac{\sin x}{x}\) equals 0, not 1.

Step-by-step solution

Identify the Function

The function given is \(\frac{\sin x}{x}\). We are looking for its behavior as \(x\) approaches \(\pi\).

Evaluate the Limit

Directly substituting \(x=\pi\) into the function, we get \(\frac{\sin \pi}{\pi}\). The \(\sin\) of \(\pi\) radians is 0, so this gives \(\frac{0}{\pi}\) which is 0, not 1 as the statement suggests.

Final Evaluation

Since the evaluated limit value (0) is not equal to what the statement suggests it is (1), we can conclude that the statement is false.

Key concepts

Trigonometric Limits

Trigonometric limits are a specific category within limits in calculus that deal with functions involving trigonometric expressions such as sine, cosine, and tangent. Understanding these limits is crucial when solving problems where trigonometric functions are at play. For example, when evaluating \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), which is a standard limit in calculus. This result is used frequently in calculus as it helps simplify complex expressions. In the exercise provided, we have a similar looking expression, \( \frac{\sin x}{x} \), but as \( x \to \pi \) instead of zero. Unlike the standard limit, this does not equal 1, since \( \sin(\pi) = 0 \). This illustrates that trigonometric limits can vary greatly depending on the point of approach and reinforce the importance of context when dealing with trigonometric functions in calculus. It highlights that knowing a few key trigonometric limits can simplify many problems.

Direct Substitution

Direct substitution is one of the simplest techniques to evaluate limits, especially helpful when the function is continuous at the point we are evaluating. It involves plugging the value directly into the function to find the limit. In the given exercise, we use direct substitution to evaluate \( \lim_{x \to \pi} \frac{\sin x}{x} \) by replacing \( x \) with \( \pi \). After substitution, we calculate \( \frac{\sin(\pi)}{\pi} = \frac{0}{\pi} \), resulting in zero.
This method is straightforward. However, it may not always work, especially if the substitution results in an undefined form, like \( \frac{0}{0} \). In such cases, you might need alternative techniques. But when you can use direct substitution, it provides a quick and efficient way to evaluate a limit. This lesson shows that while direct substitution is simple, it needs careful handling of the function's behavior at the point of interest.

Limit Evaluation Techniques

To tackle limit problems effectively, several techniques can come in handy if direct substitution fails due to indeterminate forms like \( \frac{0}{0} \). These techniques include:

• Factoring and Canceling: Often used for algebraic limits, you factor out terms to cancel them before substituting.
• Conjugates: Multiplying by a conjugate helps eliminate square roots or irrational numbers in limit problems.
• Trigonometric Identities: Utilizes identities like \( \sin^2 x + \cos^2 x = 1 \) to simplify expressions.
• L'Hôpital's Rule: Applies when dealing with indeterminate forms. It involves taking derivatives of the numerator and denominator.

Such techniques are essential when you can't directly substitute a number into a function without encountering an undefined form. They allow you to manipulate and simplify the expression to find the correct limit. Understanding when and how to apply these techniques is integral to mastering limit problems.