Problem 65 - Thomas Calculus

Question

Find the limits in Exercises \(51-66\) $$ \lim _{x \rightarrow 0^{+}} x \tan \left(\frac{\pi}{2}-x\right) $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Understanding the Function

We need to find the limit of the function \( x \tan\left(\frac{\pi}{2} - x\right) \) as \( x \to 0^+ \). When \( x \to 0^+ \), the angle \( \frac{\pi}{2} - x \) approaches \( \frac{\pi}{2} \), where the tangent function has a vertical asymptote.

Step 2: Simplify the Expression

Recall that \( \tan\left(\frac{\pi}{2} - x\right) \) is equivalent to \( \cot(x) \) due to the complementary angle identity \( \tan\left(\frac{\pi}{2} - x \right) = \cot(x) \). So, the expression becomes \( x \cdot \cot(x) = x \cdot \frac{1}{\tan(x)} = \frac{x}{\tan(x)} \).

Step 3: Evaluate the Limit

We recognize an indeterminate form \( \frac{0}{0} \) for the limit \( \lim_{x \to 0^+} \frac{x}{\tan(x)} \). Use the limit identity \( \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \) to evaluate: \( \lim_{x \to 0^+} \frac{x}{\tan(x)} = \lim_{x \to 0^+} \frac{x \cdot x}{\sin(x)} = \lim_{x \to 0^+} \frac{x^2}{\sin(x)} \).

Step 4: Use Trigonometric Limit Identity

Rearrange the expression: \( \lim_{x \to 0^+} \frac{x^2}{\sin(x)} = \lim_{x \to 0^+} x \cdot \frac{x}{\sin(x)} = \lim_{x \to 0^+} x \cdot 1 = 0 \). Here, we used \( \lim_{x \to 0^+} \frac{x}{\sin(x)} = 1 \), so the product is \( 0 \cdot 1 = 0 \).

Key Concepts

Trigonometric Limits

Trigonometric limits often involve functions such as sine, cosine, and tangent. When solving these limits, especially as the variable approaches zero or another critical point, special identities and properties are useful. In the context of the exercise, we observe that as \( x \) approaches \( 0^+ \), the angle \( \frac{\pi}{2} - x \) changes.

• The identity \( \tan\left(\frac{\pi}{2} - x\right) = \cot(x) \) is a key tool.
• Trigonometric functions can exhibit rapid changes near vertical asymptotes, such as \( \frac{\pi}{2} \).

This requires us to work carefully with their equivalent forms, allowing us to apply standard limit techniques.

Indeterminate Forms

Indeterminate forms arise frequently in calculus when limits are evaluated. These forms occur when the limit yields expressions like \( \frac{0}{0} \), \( \frac{\infty}{fty} \), or \( 0 \cdot \infty \). In the given problem, as \( x \) approaches \( 0^+ \), both the numerator \( x \) and the function \( \tan(x) \) in \( \tan\left(\frac{\pi}{2}-x\right) \) approach zero.
When we reframe \( x \tan\left(\frac{\pi}{2}-x\right) \) to \( \frac{x}{\tan(x)} \), we recognize the \( \frac{0}{0} \) form.

• This makes it an indeterminate form suitable for evaluating with known limit identities or rules.
• We can use the limit \( \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \) as a simplifying identity.

This process helps us convert the indeterminate form into a solvable expression.

Complementary Angles

The concept of complementary angles is crucial in trigonometry, particularly when working with limits involving these angles. Two angles are complementary when their sum is \( \frac{\pi}{2} \) (or 90 degrees). Trigonometric identities based on complementary angles allow us to transform expressions for easier manipulation.

• The transformation \( \tan\left(\frac{\pi}{2} - x\right) = \cot(x) \) simplifies handling the original expression.
• This identity derives from the complementary nature of the angles \( \frac{\pi}{2} \) and \( x \).

Understanding these relationships is pivotal in "flipping" trigonometric functions to their reciprocal counterparts, which are often simpler to analyze within limits.

L'Hôpital's Rule

L'Hôpital's Rule can be used to solve limits involving indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). The rule states that if the limit \( \lim_{x \to c} \frac{f(x)}{g(x)} \) is indeterminate, then \ \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \) provided the latter limit exists.
Though commonly used, in this particular exercise, we tackled the limit by using trigonometric identities instead:

• We identified that \( \frac{x}{\tan(x)} \) could be transformed using trigonometric limits.
• Directly applying L'Hôpital’s Rule was unnecessary as simpler trigonometric manipulation solved the issue.

This shows the importance of considering multiple strategies when encountering limits in calculus.