Problem 47 - Thomas Calculus in SI Units

Question

Use l'Hôpital's rule to find the limits. $$\lim _{x \rightarrow 0} \frac{x-\sin x}{x \tan x}$$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Check the indeterminate form

First, substitute the value of the limit directly into the expression: \( \lim_{x \rightarrow 0} \frac{x - \sin x}{x \tan x} \). Substituting \(x = 0\), the expression becomes \(\frac{0 - 0}{0 \cdot 0} = \frac{0}{0}\), which is an indeterminate form. Therefore, we can apply l'Hôpital's Rule.

Step 2: Differentiate the numerator and denominator

To apply l'Hôpital's Rule, differentiate the numerator \(x - \sin x\) and the denominator \(x \tan x\) with respect to \(x\). The derivative of the numerator is \(1 - \cos x\). The derivative of \(x \tan x\) is found by using the product rule: first differentiate \(x\) to get \(1\), and differentiate \(\tan x\) to get \(\sec^2 x\). So the derivative of \(x \tan x\) is \(1 \cdot \tan x + x \cdot \sec^2 x = \tan x + x \sec^2 x\).

Step 3: Apply l'Hôpital's Rule

Apply l'Hôpital's Rule to the limit expression: \(\lim_{x \rightarrow 0} \frac{1 - \cos x}{\tan x + x \sec^2 x}\). This replaces the original limit with a new limit using the derivatives calculated in Step 2.

Step 4: Evaluate the new limit

Substitute \(x = 0\) into the modified expression: \(\lim_{x \rightarrow 0} \frac{1 - \cos x}{\tan x + x \sec^2 x}\). Numerically, this becomes \(\frac{1 - 1}{0 + 0} = \frac{0}{0}\), indicating the expression remains indeterminate, and another application of l'Hôpital's Rule is necessary.

Step 5: Differentiate again

Calculate the second derivative of the numerator and the denominator. The derivative of \(1 - \cos x\) is \(\sin x\), and the derivative of \(\tan x + x \sec^2 x\) is \(\sec^2 x + (\sec^2 x + 2x \sec^2 x \tan x)\). Simplifying, it becomes \(\sec^2 x + 2x \sec^2 x \tan x\).

Step 6: Apply l'Hôpital's Rule again

Apply l'Hôpital's Rule again to \(\lim_{x \rightarrow 0} \frac{\sin x}{\sec^2 x + 2x \sec^2 x \tan x}\). Evaluate this by substituting \(x = 0\), we get \(\frac{0}{1}\), because \(\sin(0) = 0\) and \(\sec^2(0) = 1\). Therefore, the limit is \(0\).

Key Concepts

Indeterminate Forms

An indeterminate form is a mathematical expression that initially doesn’t lend itself to evaluation without more work. It's a signal that we need to employ certain techniques, like applying l'Hôpital's Rule, to resolve the ambiguity. One of the most common indeterminate forms we encounter is \( \frac{0}{0} \). This happens when both the numerator and the denominator of a limit expression approach zero simultaneously as the variable approaches a particular value. In our exercise, substituting \( x = 0 \) in the limit expression \( \frac{x - \sin x}{x \tan x} \) results in the form \( \frac{0}{0} \). This suggests that direct evaluation won't work, and we should apply more advanced methods to solve it. Recognizing these forms not only helps in limits but also paves the way to understanding more complex calculus concepts, such as series and integral calculation.

Differentiation

Differentiation is a core concept in calculus that helps us find how a function changes at any given point. It's the first step in applying l'Hôpital's Rule to resolve indeterminate forms. In this exercise, to break down the limit, we differentiated both the numerator and the denominator. The numerator \( x - \sin x \) was differentiated to yield \( 1 - \cos x \).

• Product Rule: To differentiate the denominator \( x \tan x \), we used the product rule. This rule states that to differentiate two multiplied functions, we find \( f'(x)g(x) + f(x)g'(x) \). Here, \( x \) is one function and \( \tan x \) is the other.
• As a result, the derivative of the denominator is \( \tan x + x \sec^2 x \), where \( \sec^2 x \) is the derivative of \( \tan x \).

Differentiation not only helps in resolving indeterminate forms in limits but is also vital in various fields like physics and economics, wherever changes and slopes are analyzed.

Calculus Limits

Calculus limits allow us to understand the behavior of functions as the input approaches a certain value. They are fundamental to calculus, bridging derivatives and integrals together. Limits help in identifying the point of convergence or divergence of a series or a function. In our instance, the task was to solve \( \lim_{x \rightarrow 0} \frac{x - \sin x}{x \tan x} \).

• When expressions result in forms like \( \frac{0}{0} \), calculus limits, combined with tools like l'Hôpital's Rule, become essential in finding meaningful results.
• Repeated application of l'Hôpital's Rule in the exercise involved differentiating again after identifying that the limit \( \frac{0}{0} \) persisted even after initial differentiation.
• This process eventually led us to \( \lim_{x \rightarrow 0} \frac{\sin x}{\sec^2 x + 2x \sec^2 x \tan x} \), which simplified to \( 0 \), demonstrating how calculus limits can elucidate particular behaviors of functions at critical points.

By mastering limits, one gains a deeper understanding of continuity, approaching infinity, and the precise definition of derivatives.