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What happens if you try to use I'Hospital's Rule to find the limit? Evaluate the limit using another method. \(\lim _{x \rightarrow(\pi / 2)^{-}} \frac{\sec x}{\tan x}\)

Short Answer

Expert verified
I'Hospital's Rule is not applicable; the limit is \(+\infty\).

Step by step solution

01

Understand I'Hospital's Rule

I'Hospital's Rule can be applied to limits that result in indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). We first need to identify if our limit results in one of these forms when \( x \to \left( \frac{\pi}{2} \right)^{-} \).
02

Find the Original Limit Behavior

Rewrite \( \sec x \) and \( \tan x \) in terms of \( \sin x \) and \( \cos x \):\[ \sec x = \frac{1}{\cos x}, \quad \tan x = \frac{\sin x}{\cos x} \].Therefore, the limit becomes:\[ \lim_{x \to (\pi/2)^-} \frac{1/\cos x}{\sin x/\cos x} = \lim_{x \to (\pi/2)^-} \frac{1}{\sin x} \].Evaluate this behavior as \( x \rightarrow (\pi / 2)^{-} \).
03

Evaluate the Limit Without I'Hospital's Rule

As \( x \to (\pi/2)^- \), \( \sin x \to 1 \) but since we're approaching \( \pi/2 \) from the left, \( \sin x \to 0^+ \), leading to \( \frac{1}{\sin x} \rightarrow +\infty \). Thus the limit is \(+\infty \).
04

Confirm Indeterminate Form for I'Hospital's Rule

Since the form of \( \frac{1/\cos x}{\sin x/\cos x} \) simplifies to \( \frac{1}{\sin x} \), it doesn't evaluate to an indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) directly. Therefore, I'Hospital's Rule is not applicable.
05

Conclude the Correct Evaluation Method

I'Hospital's Rule is not applicable for this limit. Instead, by direct substitution and consideration of the behavior of \( \sin x \) as \( x \to (\pi/2)^- \), the limit is evaluated to be \(+\infty\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

L'Hospital's Rule
L'Hospital's Rule is a helpful tool in calculus for finding limits that result in indeterminate forms. The rule is applicable when a limit initially results in forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These are special cases where direct substitution into the function doesn’t give us a clear answer. When you encounter such a form, you can differentiate the numerator and the denominator separately and then attempt the limit again. This rule simplifies the problem significantly and often allows for limits to be calculated that would be otherwise very difficult.In the problem at hand, it's crucial to first identify the form of the limit before deciding on the use of L'Hospital’s Rule. If the limit simplifies directly to a determinate form, such as in this case \( \frac{1}{0^+} \rightarrow +\infty \), the rule is not needed. This concept highlights why understanding the behavior of the trigonometric limits and the function as we approach certain values is as important as knowing when to apply calculus rules like L'Hospital's Rule.
Indeterminate Forms
Indeterminate forms are expressions in calculus that don’t initially provide us with a straightforward limit value. They are special cases that require careful analysis to resolve the limits correctly. Some of the common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \times \infty \), among others.Here is where L'Hospital's Rule often becomes relevant. Once we identify that a function approaches one of these forms, we can turn to this rule to differentiate each part of the function before re-evaluating the limit. However, it's important not to misuse this rule, as it only works on specific forms, mainly \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \). In the given exercise, the original function \( \frac{\sec x}{\tan x} \) transformed to \( \frac{1}{\sin x} \) leading it directly to \( \frac{1}{0^+} \), which is determinate and shows approaching \(+\infty\), meaning it wasn't necessary to invoke L'Hospital’s Rule.Ensuring you first rewrite or analyze the limit thoroughly helps clarify whether you are indeed encountering an indeterminate form.
Trigonometric Limits
Trigonometric limits are a common focus within calculus, often requiring special techniques for evaluating. They involve functions such as \( \sin x \), \( \cos x \), \( \tan x \), and related functions. Calculating these limits usually involves understanding the behavior of these functions as they approach significant points in their domain.In the exercise, the limit involves \( \frac{\sec x}{\tan x} \). By rewriting these functions using \( \sin x \) and \( \cos x \), the problem becomes more manageable. Using the identities:
  • \( \sec x = \frac{1}{\cos x} \)
  • \( \tan x = \frac{\sin x}{\cos x} \)
helps to simplify the expression to \( \frac{1}{\sin x} \), a form where behavior at specific points, like \( x \to (\pi/2)^- \), can be easily evaluated. Focusing on the trigonometric behavior helps understand why, as \( x \) approaches \( \pi/2 \), \( \sin x \) approaches \( 0 \), ultimately leading us to the conclusion that the limit approaches \(+\infty\). This kind of intuitive evaluation plays a significant role in correctly interpreting and solving trigonometric limits.

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Most popular questions from this chapter

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