Problem 26 - Thomas Calculus in SI Units

Question

Use l'Hôpital's rule to find the limits. $$\lim _{x \rightarrow(\pi / 2)^{-}}\left(\frac{\pi}{2}-x\right) \tan x$$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Identify Indeterminate Form

First, substitute \(x = \frac{\pi}{2}^{-}\) into the function \((\frac{\pi}{2} - x) \tan{x}\) to check if it's an indeterminate form. As \(x ightarrow \frac{\pi}{2}^{-}, \\tan{x} ightarrow \infty\) and \(\frac{\pi}{2} - x ightarrow 0^{+}\).This multiplication of \(0\cdot \infty\) is an indeterminate form.

Step 2: Expression Conversion

Convert the original form \(\left(\frac{\pi}{2} - x\right) \tan{x}\) to a fraction suitable for l'Hôpital's rule. This can be rewritten as \(\frac{\frac{\pi}{2} - x}{\cot{x}}\). Now, \(\frac{\pi}{2} - x ightarrow 0\) and \(\cot{x} = \frac{1}{\tan{x}} \rightarrow 0\) as \(x \rightarrow \frac{\pi}{2}\). This is a \frac{0}{0}\ form, so l'Hôpital's rule can be applied.

Step 3: Apply l'Hôpital's Rule

Differentiate the numerator and denominator separately.The derivative of \(\frac{\pi}{2} - x\) with respect to \(x\) is \(-1\).The derivative of \(\cot{x}\) with respect to \(x\) is \(-\csc^2{x}\).Thus, the limit becomes \(\lim_{x \rightarrow \frac{\pi}{2}^-} \frac{-1}{-\csc^2{x}} = \lim_{x \rightarrow \frac{\pi}{2}^-} \frac{1}{\csc^2{x}} \).

Step 4: Evaluate the Final Limit

Simplify \(\frac{1}{\csc^2{x}} = \sin^2{x}\).As \(x \rightarrow \frac{\pi}{2}^-\), \(\sin{x} \rightarrow 1\), so \(\sin^2{x} \rightarrow 1\).Thus, the final limit is \(1\).

Key Concepts

Understanding Indeterminate Forms

When dealing with limits, encountering an indeterminate form can make things tricky. These forms occur when a direct substitution in the limit expression results in an undefined or ambiguous answer. Common examples include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), and others. They indicate that we can't simply plug in the values and need a deeper analysis. In this problem, when trying to find the limit \( \lim _{x \rightarrow(\pi / 2)^{-}}\left(\frac{\pi}{2}-x\right) \tan x \), substituting the values gives us \( 0 \cdot \infty \), which is indeterminate. The indeterminate nature means that the limit may exist but requires a careful method to evaluate. This usually means rewriting the expression or using special techniques such as l'Hôpital's rule.

Limit Evaluation Techniques

Evaluating limits of expressions that lead to indeterminate forms often requires strategic manipulation of the expression. For instance, if you encounter a \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) form, you can use l'Hôpital's rule. This rule involves differentiating the numerator and denominator until a determined limit is achieved. In our problem, the expression \( (\frac{\pi}{2} - x) \tan{x} \) was transformed into \( \frac{\frac{\pi}{2} - x}{\cot{x}} \). This changes a multiplication form into a fraction, which can be tackled with l'Hôpital's rule. By differentiating the new top and bottom separately, the limit becomes manageable: \( \lim_{x \rightarrow \frac{\pi}{2}^-} \frac{-1}{-\csc^2{x}} = \lim_{x \rightarrow \frac{\pi}{2}} \sin^2{x} \). After applying derivatives, many limits become straightforward to compute.

Trigonometric Limits and Their Behavior

Trigonometric functions have specific behaviors near certain key points, which often play a crucial role in limit evaluation. In the exercise, as \( x \) approaches \( \frac{\pi}{2}^- \), \( \tan{x} \) approaches \( \infty \), and \( \sin{x} \) approaches 1. This behavior is leveraged during the evaluation of limits involving trigonometric terms. By knowing how these functions act, we can convert seemingly complex terms into simpler forms. For example, after applying l'Hôpital's rule, the problem's expression simplifies to \( \sin^2{x} \). Understanding these trigonometric limits lets us conclude that \( \lim_{x \rightarrow \frac{\pi}{2}^-} \sin^2{x} = 1 \). Thus, knowing trigonometric function behaviors helps break down limits into solvable components.