Question

, use a calculator to find the indicated limit. Use a graphing calculator to plot the function near the limit point. $$ \lim _{x \rightarrow \pi} \frac{1+\sin (x-3 \pi / 2)}{x-\pi} $$

Short answer

The limit is 0.

Step-by-step solution

Understanding the Problem

We need to find the limit as \( x \to \pi \) of the function \( f(x) = \frac{1+\sin(x-3\pi/2)}{x-\pi} \). This involves evaluating the value the function approaches as \( x \) gets very close to \( \pi \).

Consider the Function

The given function is \( \frac{1+\sin(x-3\pi/2)}{x-\pi} \). Notice that as \( x \to \pi \), the denominator \( x-\pi \) approaches zero, which might suggest a potential division by zero issue unless the numerator approaches zero as well.

Substitute and Simplify the Numerator

Substitute \( x = \pi \) into the numerator \( 1 + \sin(x-3\pi/2) \). Calculate \( \sin(\pi - 3\pi/2) = \sin(-\pi/2) = -1 \). Thus, the numerator becomes \( 1 - 1 = 0 \).

Identify the Indeterminate Form

Both the numerator and the denominator approach zero as \( x \rightarrow \pi \), creating a \( 0/0 \) indeterminate form. This suggests that L'Hôpital's rule might be applicable.

Apply L'Hôpital's Rule

L'Hôpital's Rule states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then you can find it by \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \) if the limit exists. Differentiate the numerator and the denominator. The derivative of the numerator \( 1+\sin(x-3\pi/2) \) is \( \cos(x-3\pi/2) \). The derivative of the denominator \( x-\pi \) is \( 1 \).

Evaluate the Limit Using the Derivatives

Now, find the limit \( \lim_{x \to \pi} \frac{\cos(x-3\pi/2)}{1} = \lim_{x \to \pi} \cos(x-3\pi/2) \). Substitute \( x = \pi \) into \( \cos(x-3\pi/2) \) to get \( \cos(\pi - 3\pi/2) = \cos(-\pi/2) = 0 \).

Conclusion

Hence, the limit is \( 0 \). We used a graphing calculator to confirm this result by observing the graph of the function near \( x = \pi \), which also approaches 0.

Key concepts

L'Hôpital's Rule

One of the powerful tools in calculus, especially when dealing with limits that result in indeterminate forms like \( \frac{0}{0} \), is L'Hôpital's Rule. This rule provides a way to simplify such limits by differentiating the numerator and the denominator separately. Here's how it works: if you have a function \( \lim_{x \to c} \frac{f(x)}{g(x)} \), and it results in an indeterminate form of \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), you can differentiate the top \( f'(x) \) and the bottom \( g'(x) \) and then find the limit of this new fraction.

It is crucial, however, that both \( f' \) and \( g' \) exist and the limit resulting from \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \) exists as well. If so, this can simplify complex problems significantly and give you the correct answer. In our specific problem, both the numerator and denominator reach zero at \( x = \pi \), indicating L'Hôpital's Rule is applicable.

Indeterminate Forms

Indeterminate forms occur in calculus when the expression of a limit cannot directly suggest the limit value. Key forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \times \infty \), and similar. These are problematic because they don't provide clear information on the behavior of the fraction as \( x \) approaches a certain value.

In this exercise, as \( x \to \pi \), the given limit \( \lim \frac{1+\sin(x-3\pi/2)}{x-\pi} \) transformed into the indeterminate form of \( \frac{0}{0} \). This signals that straightforward substitution isn't feasible, prompting the need to find alternative approaches like L'Hôpital's Rule.

Understanding indeterminate forms can highlight places in a problem where calculus tools like L'Hôpital's Rule can be instrumental.

Trigonometric Limits

Trigonometric limits involve functions containing trigonometric components like sine and cosine. These functions often result in indeterminate forms due to their periodic nature, requiring special attention to angles and transformations.

In our problem, the term \( \sin(x-3\pi/2) \) is integral. By substituting \( x = \pi \), we evaluated \( \sin(\pi - 3\pi/2) \), which simplifies to \( \sin(-\pi/2) \). Here, understanding the sine wave is crucial, as it shows the trigonometric function approaching zero, contributing to the indeterminate form.

Tackling trigonometric limits often involves utilizing identities, periodic properties, and transformations to simplify a limit problem into an easily evaluable form.

Graphing Calculator

A graphing calculator can provide visual insights into calculus problems by displaying the behavior of a function near a point of interest. For limits, this visual representation is crucial to observe how the function behaves as it approaches the limit.

In our exercise, sketching the function \( \frac{1+\sin(x-3\pi/2)}{x-\pi} \) near \( x = \pi \) confirms the calculated limit value. As we move closer to \( \pi \), the function trends towards zero, echoing the computed result using L'Hôpital's Rule.

• It helps visualize complex behaviors of functions around discontinuities or indeterminate forms.
• Provides a check against analytical solutions by graphing the approach of \( x \to c \).

By combining analytical techniques and graphical insights, students can more thoroughly understand limits and their computations.