Problem 53 - Calculus Early Transcendentals

Question

Evaluate the following limits. $$\lim _{\theta \rightarrow \pi / 2^{-}}(\tan \theta-\sec \theta)$$

Step-by-Step Solution

Answer

Question: Evaluate the limit of the expression \(\tan \theta - \sec \theta\) as \(\theta\) approaches \(\frac{\pi}{2}\) from the left. Answer: The limit of the expression \(\tan \theta - \sec \theta\) as \(\theta\) approaches \(\frac{\pi}{2}\) from the left is 0.

Step 1: Recall the Definitions of Tangent and Secant

The tangent function is given by \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), and the secant function is given by \(\sec \theta = \frac{1}{\cos \theta}\). Our expression then becomes: $$\tan \theta - \sec \theta = \frac{\sin \theta}{\cos \theta} - \frac{1}{\cos \theta}$$

Step 2: Find a Common Denominator

In order to subtract the fractions, we need to find a common denominator. In this case, our common denominator will be \(\cos \theta\). $$\frac{\sin \theta - 1}{\cos \theta}$$

Step 3: Evaluate the Limit

Now, we want to evaluate the limit as \(\theta\) approaches \(\frac{\pi}{2}\) from the left: $$\lim _{\theta \rightarrow \frac{\pi}{2}^-}\left(\frac{\sin \theta - 1}{\cos \theta}\right)$$ When approaching from the left, \(\cos \theta\) approaches 0, but the sine function approaches 1. Thus, the limit can be written as: $$\lim _{\theta \rightarrow \frac{\pi}{2}^-}\left(\frac{\sin \theta - 1}{\cos \theta}\right) = \frac{\lim _{\theta \rightarrow \frac{\pi}{2}^-} (\sin \theta - 1)}{\lim _{\theta \rightarrow \frac{\pi}{2}^-} (\cos \theta)}$$ Evaluating each of these individual limits, we have: $$\lim _{\theta \rightarrow \frac{\pi}{2}^-} (\sin \theta - 1) = 1 - 1 = 0$$ and $$\lim _{\theta \rightarrow \frac{\pi}{2}^-} (\cos \theta) = 0$$

Step 4: Apply L'Hôpital's Rule

Since both the numerator and denominator of the expression tend to 0, we can apply L'Hôpital's Rule. To apply L'Hôpital's Rule, we differentiate the numerator and denominator separately with respect to \(\theta\): $$\frac{d(\sin \theta - 1)}{d\theta} = \cos \theta$$ $$\frac{d(\cos \theta)}{d\theta} = -\sin \theta$$ Now, we can re-evaluate the limit of the expression as \(\theta\) approaches \(\frac{\pi}{2}\) from the left: $$\lim _{\theta \rightarrow \frac{\pi}{2}^-}\left(\frac{\cos \theta}{-\sin \theta}\right)$$ As before, as \(\theta\) approaches \(\frac{\pi}{2}\) from the left, \(\cos \theta\) approaches 0 and \(\sin \theta\) approaches 1: $$\lim _{\theta \rightarrow \frac{\pi}{2}^-}\left(\frac{\cos \theta}{-\sin \theta}\right) = \frac{0}{-1} = 0$$ So the limit of the expression is 0.

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a useful method in calculus to find limits of indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When a direct substitution in a limit gives you one of these forms, you can apply L'Hôpital's Rule. This rule states that you can take the derivative of the numerator and the derivative of the denominator separately, then calculate the limit again.
In our example, when evaluating \( \lim _{\theta \rightarrow \frac{\pi}{2}^-}\left(\frac{\sin \theta - 1}{\cos \theta}\right) \), both the numerator and denominator approach 0 as \( \theta \) approaches \( \frac{\pi}{2} \) from the left. This forms a \( \frac{0}{0} \) indeterminate form. By differentiating the numerator to get \( \cos \theta \) and the denominator to get \( -\sin \theta \), we can apply L'Hôpital's Rule again resulting in:

• The new expression \( \lim _{\theta \rightarrow \frac{\pi}{2}^-}\left(\frac{\cos \theta}{-\sin \theta}\right) \).
• As \( \theta \to \frac{\pi}{2}^- \), \( \cos \theta \to 0 \) and \( \sin \theta \to 1 \), resulting in \( \frac{0}{-1} = 0 \).

L'Hôpital's Rule helps us resolve the original limit to 0 by transforming the expression into a solvable form.

Trigonometric Limits

Trigonometric limits involve evaluating limits that include trigonometric functions such as sine, cosine, and tangent. They often require special consideration because of their behavior near particular angles and their periodic nature.
When approaching trigonometric limits, it's important to know key angles such as \( 0, \frac{\pi}{2}, \pi, \) etc., where trigonometric functions take on predictable values. For example, \( \lim_{\theta \to 0}\sin\theta = 0 \) and \( \lim_{\theta \to \frac{\pi}{2}^-}\cos\theta = 0 \).

• In the primary example, the term \( \tan \theta - \sec \theta \) is handled by converting it to \( \frac{\sin \theta}{\cos \theta} - \frac{1}{\cos \theta} \), and simplifying to \( \frac{\sin\theta - 1}{\cos \theta} \).
• As \( \theta \to \frac{\pi}{2}^- \), \( \sin \theta \to 1 \) and \( \cos \theta \to 0 \) leads to a limit in the form of \( \frac{0}{0} \).

This case shows the complexity and beauty of trigonometric limits and how they often mesh with other calculus concepts like L'Hôpital's Rule.

Function Behavior at Infinity

Understanding how functions behave as they approach infinity is crucial in calculus, particularly when dealing with limits, derivatives, and integrals. At infinity, many functions display asymptotic behavior, meaning they approach some limit.
When analyzing function behavior at or approaching certain critical points, such as \( \theta \to \frac{\pi}{2}^- \), which isn't infinity but shares characteristics due to a singularity, the function values can become extreme or undefined. In these cases:

• Functions like tangent or secant can approach infinity or negative infinity, depending on direction and angle.
• Trigonometric functions may exhibit discontinuities or zeros that vary between \( -1 \) and \( 1 \), making precise limit calculations essential.

In the original problem, even though we aren't strictly approaching infinity, the concept of function behavior around critical points like \( \frac{\pi}{2}^- \) helps us analyze the precise behavior of the function, ensuring comprehensive understanding of limits involving trigonometric functions.