Problem 49 - Thomas Calculus

Question

Use l'Hopital's rule to find the limits in Exercises \(7-50\) . $$ \lim _{\theta \rightarrow 0} \frac{\theta-\sin \theta \cos \theta}{\tan \theta-\theta} $$

Step-by-Step Solution

Answer

The limit is 2.

Step 1: Identify the Indeterminate Form

First, we need to determine whether the given limit is in an indeterminate form. Substitute \( \theta = 0 \) in the expression: \[ \frac{0 - \sin 0 \cdot \cos 0}{\tan 0 - 0} = \frac{0}{0}. \]Since this is a \( \frac{0}{0} \) indeterminate form, we can use l'Hôpital's rule.

Step 2: Apply l'Hôpital's Rule

l'Hôpital's Rule states that if a limit results in the indeterminate form \( \frac{0}{0} \), the limit of the ratio of the derivatives of the functions can be used instead. Therefore, differentiate the numerator \( f(\theta) = \theta - \sin \theta \cos \theta \) and the denominator \( g(\theta) = \tan \theta - \theta \).

Step 3: Differentiate the Numerator

The numerator is \( f(\theta) = \theta - \sin \theta \cos \theta \). Apply the product rule to \( \sin \theta \cos \theta \):\[ f'(\theta) = 1 - \left( \cos \theta \cdot \cos \theta - \sin \theta \cdot \sin \theta \right) = 1 - (\cos^2 \theta - \sin^2 \theta). \]Simplify to \[ f'(\theta) = 1 - \cos 2\theta. \]

Step 4: Differentiate the Denominator

The denominator is \( g(\theta) = \tan \theta - \theta \). The derivative of \( \tan \theta \) is \( \sec^2 \theta \) and of \( \theta \) is 1:\[ g'(\theta) = \sec^2 \theta - 1. \]

Step 5: Compute the Limit of the Derivatives

Now, evaluate the limit using the derivatives:\[ \lim_{\theta \to 0} \frac{f'(\theta)}{g'(\theta)} = \lim_{\theta \to 0} \frac{1 - \cos 2\theta}{\sec^2 \theta - 1}. \]Substitute \( \theta = 0 \):\[ \frac{1 - \cos 0}{\sec^2 0 - 1} = \frac{1 - 1}{1 - 1}. \]This also gives \( \frac{0}{0} \), so we need to simplify further. Apply l'Hôpital's rule again if needed.

Step 6: Simplify Further by L'Hôpital's Rule Again

We need to use l'Hôpital's rule once more. Differentiate:\[ f''(\theta) = 2\sin(2\theta), \]\[ g''(\theta) = 2\sec^2\theta \tan\theta. \]Evaluate the limit:\[ \lim_{\theta \to 0} \frac{2\sin(2\theta)}{2\sec^2\theta \tan\theta} = \lim_{\theta \to 0} \frac{\sin(2\theta)}{\sec^2\theta \tan\theta}. \]As \( \theta \rightarrow 0 \), this becomes \( \lim_{\theta \to 0} \frac{2\theta}{\theta} = 2 \). Hence, the limit is 2.

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms are expressions that are not immediately determinable when substituting specific values. A common example is the \( \frac{0}{0} \) form, where both the numerator and the denominator approach zero. Such forms suggest that we can apply further mathematical techniques, such as l'Hopital's Rule, to evaluate limits. Determining whether an expression is indeterminate is crucial before proceeding with any method, including trigonometric and other complex functions.

• Checks if the limit seems undefined directly.
• Indicates the need for additional techniques like differentiation.

Once identified, indeterminate forms require manipulation to simplify or transform the expression into a determinate one, revealing the true limit of the function as the variable approaches the given value.

Trigonometric Limits

Trigonometric limits involve finding the limit of functions that contain trigonometric expressions such as \(\sin \theta\) or \(\cos \theta\). These often pose specific challenges due to periodicity and range. When working with trigonometric limits, it is essential to understand fundamental identities:

• \( \sin^2 \theta + \cos^2 \theta = 1 \)
• Other identities like \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)

In our example, trigonometric identities and properties of these functions at significant points, like \(\theta = 0\), help us understand behavior at approaching zeros. Knowing these can help analyze and simplify complex expressions, getting us closer to accurately finding the limit.

Calculus Limits

Limits in calculus are fundamental in understanding the behavior of functions as variables approach specific values. These are the foundation for calculus concepts like continuity and derivatives. Calculus limits help:

• Analyze the behavior of functions as they near a point.
• Determine function stability and potential discontinuities.

When evaluating a limit, especially one involving complex functions like trigonometric polynomials or involving indeterminate forms, calculating limits helps to analyze what the expression tends toward as the input approaches a certain value.

Derivatives

Derivatives, essential in calculus, represent the rate of change or slope of a function. When dealing with the calculation of limits in indeterminate forms, derivatives allow us to apply l'Hopital's Rule effectively. For instance, we take the derivatives of the numerator and denominator separately:

• Using product and chain rules for complex expressions.
• Simplifying expressions to analyze limit behavior.

In solving our exercise, we differentiated both the top and bottom parts of the fraction until further indeterminate forms are resolved. Understanding derivatives is crucial for interpreting and simplifying complex limit problems, thus revealing not just the limit, but insights into the function's behavior near that limit.