Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find each limit. Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$ \lim _{x \rightarrow 0}\left(\csc x-\frac{1}{x}\right) $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Check the Form

To check if we can apply l'Hôpital's Rule, substitute \(x = 0\) into \(f(x) = \csc(x) - \frac{1}{x}\). This gives us \(\csc(0) = \frac{1}{\sin(0)} = \frac{1}{0}\) which is undefined, and \(\frac{1}{0}\) is also undefined, resulting in the indeterminate form \(\infty - \infty\). We cannot apply l'Hôpital's Rule directly in this form. Instead, we need to convert this form to \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) to apply l'Hôpital's Rule.

Step 2: Rewrite the Expression

To convert \(\csc(x) - \frac{1}{x}\) into a form suitable for l'Hôpital's Rule, find a common denominator:\[\csc(x) - \frac{1}{x} = \frac{1}{\sin(x)} - \frac{1}{x} = \frac{x - \sin(x)}{x\sin(x)}\].Now, we have the form \(\frac{0}{0}\) as \(x \rightarrow 0\).

Step 3: Apply l'Hôpital's Rule

Since we have \(\frac{0}{0}\), we apply l'Hôpital’s Rule by differentiating the numerator and the denominator:Numerator derivative:\[ \frac{d}{dx}[x - \sin(x)] = 1 - \cos(x) \]Denominator derivative:\[ \frac{d}{dx}[x \sin(x)] = \sin(x) + x \cos(x) \]Apply l’Hôpital’s Rule:\[\lim_{x \to 0} \frac{x - \sin(x)}{x\sin(x)} = \lim_{x \to 0} \frac{1 - \cos(x)}{\sin(x) + x\cos(x)}\].

Step 4: Evaluate the Limit

Evaluate the limit as \(x \rightarrow 0\):The numerator \(1 - \cos(x)\) approaches 0, and the denominator \(\sin(x) + x\cos(x)\) also approaches 0, forming \(\frac{0}{0}\). We apply l'Hôpital's Rule again.Differentiate the numerator:\[\frac{d}{dx}[1 - \cos(x)] = \sin(x)\]Differentiate the denominator:\[\frac{d}{dx}[\sin(x) + x\cos(x)] = \cos(x) - x\sin(x) + \sin(x)\]The new limit is:\[\lim_{x \to 0} \frac{\sin(x)}{\cos(x) - x\sin(x) + \sin(x)}\].Since both the numerator and denominator tend to 0 as \(x \to 0\), we find:\(\sin(x) \approx x, \quad \text{and} \quad \cos(x) \approx 1 \quad \text{around zero.}\)\[\lim_{x \to 0} \frac{x}{1} = 0\].

Step 5: Conclude with the Limit

The limit after applying l'Hôpital's Rule and evaluating is:\[\lim_{x \to 0} \left(\csc(x) - \frac{1}{x}\right) = 0\].This solves the indeterminate form after appropriate manipulations.

Key Concepts

l'Hôpital's Rule

l'Hôpital's Rule is a very useful tool in calculus for evaluating limits that result in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When you encounter these indeterminate forms, it means that if you plug in the value, the function becomes undefined or ambiguous.
However, if you can rewrite the limit as a quotient such that both the numerator and the denominator approach zero or infinity, you can apply l'Hôpital's Rule. This involves differentiating the numerator and the denominator separately.

Before using l'Hôpital's Rule, make sure:

• To verify that substituting into the original function does produce an indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
• That the functions you’re working with have derivatives at the point in question.

This method simplifies the process of dealing with functions that are otherwise tricky to handle near certain points.
In essence, it lets you take a complicated limit and reduces it to a problem of finding simpler derivatives.

Indeterminate Forms

Indeterminate forms arise in calculus when the limit of a function does not lead to a precise answer due to its form.
Here are some key types of indeterminate forms:

• \( \frac{0}{0} \)
• \( \frac{\infty}{\infty} \)
• \( \infty - \infty \)

In the problem we discussed, we initially faced the form \( \infty - \infty \) which is not directly suitable for l'Hôpital's Rule. However, indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) can be handled using this rule.
The idea is to manipulate the function to transform it into a suitable form for the rule. These transformations involve algebraic tricks, like finding a common denominator, to change it into something like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). Once accomplished, l'Hôpital's Rule can be applied to evaluate the limit correctly.
Understanding and identifying the correct indeterminate form is crucial to solving limits effectively.

Trigonometric Functions

Trigonometric functions like \( \sin(x) \), \( \cos(x) \), and \( \csc(x) \) have their own unique behaviors and identities, especially as their input values approach certain limits like zero.
For the given exercise, the trigonometric function \( \csc(x) \) is the reciprocal of \( \sin(x) \). This inherently makes it undefined at points where \( \sin(x) = 0 \), such as when \( x = 0 \).
The key to solving limits involving trigonometric functions often lies in understanding their standard limits and approximations. For instance, near zero, \( \sin(x) \approx x \) and \( \cos(x) \approx 1 \).

• Remembering these approximations can help greatly simplify calculations.
• Recognizing identities and their derivatives is crucial when applying l'Hôpital's Rule.

In limit problems, especially those that concern small angles or values near zero, using these trigonometric identities and approximations helps in transforming complex expressions into simpler ones, aiding in evaluating the limit accurately.