Problem 11 - Calculus - AP Edition

Question

Find the limits. $$\lim _{x \rightarrow \pi^{+}} \frac{\sin x}{x-\pi}$$

Step-by-Step Solution

Answer

The limit is -1.

Step 1: Identify Form of the Limit

Let's evaluate the given limit \(\lim_{x \to \pi^{+}} \frac{\sin x}{x - \pi}\). As \(x\) approaches \(\pi\) from the right (denoted by \(+\)), the value of \(x - \pi\) approaches 0 from the positive side. Since \(\sin \pi = 0\), the expression takes the form \(\frac{0}{0^{+}}\), a type of indeterminate form.

Step 2: Utilize L'Hôpital's Rule

Since the limit is of the form \(\frac{0}{0^{+}}\), we can apply L'Hôpital's Rule. This rule states that if the limit \(\frac{f(x)}{g(x)}\) yields \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then \(\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}\), provided that the limit on the right exists.

Step 3: Differentiate Numerator and Denominator

Differentiate the numerator \(\sin x\) to obtain \(\cos x\). Differentiate the denominator \(x - \pi\) to obtain 1. The original limit is transformed to \(\lim_{x \to \pi^{+}} \frac{\cos x}{1}\).

Step 4: Evaluate the New Limit

To evaluate \(\lim_{x \to \pi^{+}} \cos x\), observe that \(\cos x\) is continuous and \(\cos \pi = -1\). Therefore, the limit simplifies to \(-1\).

Key Concepts

Indeterminate Forms

When we attempt to solve the limit \( \lim_{x \to \pi^{+}} \frac{\sin x}{x - \pi} \), we're faced with an indeterminate form. Indeterminate forms, like \( \frac{0}{0} \), happen when the direct substitution of values in a function leads to a situation where the outcome is not immediately clear. Indeterminate forms mean there's more work to do. Instead of attaining a single value, it suggests a more complex structure where multiple possible outcomes exist.

These forms include:

• \( \frac{0}{0} \)
• \( \frac{\infty}{\infty} \)
• \( 0 \cdot \infty \)
• \( \infty - \infty \)
• \( 0^0 \)
• \( 1^\infty \)
• \( \infty^0 \)

In our case, as \(x\) approaches \(\pi\) from the right, we are looking at \(\frac{0}{0^{+}}\). This inspires the use of specific techniques, like L'Hôpital's Rule, to evaluate limits that initially provide ambiguous results.

L'Hôpital's Rule

Faced with an indeterminate form like \( \frac{0}{0} \), L'Hôpital's Rule becomes a powerful tool. This rule is extremely useful when finding limits that result in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) after direct substitution.

L'Hôpital's Rule states that for functions \( f(x) \) and \( g(x) \), if the limit \( \frac{f(x)}{g(x)} \) yields an indeterminate form as \( x \) approaches some value \( c \), then we can differentiate the numerator and denominator separately. The limit becomes \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \), provided this new limit exists.

Let's apply this to our example:

• Differentiate \( \sin x \) to get \( \cos x \).
• Differentiate \( x - \pi \) to get \( 1 \).

Now, the problem reduces to finding \( \lim_{x \to \pi^{+}} \frac{\cos x}{1} \), which allows us to solve the limit without the complications of the original indeterminate form.

Trigonometric Limits

Trigonometric limits form an important component of calculus, especially when dealing with functions like sine and cosine. The continuity properties of these functions often make them easier to manage once transformed.

In the example of \( \lim_{x \to \pi^{+}} \cos x \), we can easily evaluate this because cosine is a continuous function. As \( x \) approaches \( \pi \), \( \cos x \) approaches \( \cos \pi \). Known values of trigonometric functions are key to solving these limits quickly.

Here are some essential trigonometric limits:

• \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)
• \( \cos(0) = 1 \)
• \( \sin(\pi) = 0 \)
• \( \cos(\pi) = -1 \)

By recognizing these patterns and leveraging known values and limits, we can simplify our calculations tremendously, as seen in transforming our original problem into evaluating \( \lim_{x \to \pi^{+}} \cos x \), which directly leads to \( -1 \).