Problem 26 - Thomas Calculus

Question

Find the limits in Exercises \(21-42\) $$\lim _{t, 0} \frac{2 t}{\tan t}$$

Step-by-Step Solution

Answer

2

Step 1: Identify Limit Type

We need to find \( \lim _{t \to 0} \frac{2t}{\tan t} \). Notice that as \( t \to 0 \), both the numerator \( 2t \) and the denominator \( \tan t \) approach 0. This is an indeterminate form \( \frac{0}{0} \), so we can apply L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule

L'Hôpital's Rule states that if you have an indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then \[ \lim _{x \to a} \frac{f(x)}{g(x)} = \lim _{x \to a} \frac{f'(x)}{g'(x)} \] assuming the limit on the right exists. Here \( f(t) = 2t \) and \( g(t) = \tan t \).

Step 3: Differentiate Numerator and Denominator

Differentiate the numerator \( f(t) = 2t \) to get \( f'(t) = 2 \). Differentiate the denominator \( g(t) = \tan t \) to get \( g'(t) = \sec^2 t \).

Step 4: Substitute the Derivatives into the Limit

According to L'Hôpital's Rule, substitute the derivatives into the limit: \[ \lim _{t \to 0} \frac{f'(t)}{g'(t)} = \lim _{t \to 0} \frac{2}{\sec^2 t} \].

Step 5: Simplify the Expression

Recall that \( \sec t = \frac{1}{\cos t} \), so \( \sec^2 t = \frac{1}{\cos^2 t} \). Thus, \[ \frac{2}{\sec^2 t} = 2 \cdot \cos^2 t \].

Step 6: Evaluate the Limit

Evaluate \( \lim _{t \to 0} 2 \cos^2 t \). As \( t \to 0 \), \( \cos t \to 1 \), so \( \cos^2 t \to 1^2 = 1 \). Therefore, \[ \lim _{t \to 0} 2 \cos^2 t = 2 \].

Key Concepts

Indeterminate Forms

In calculus, the concept of indeterminate forms arises in the study of limits. When attempting to evaluate certain limits, you may find expressions of the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These expressions are known as indeterminate forms. Why? Because just looking at the form doesn't tell us what the limit actually is—it could be anything!
\( \frac{0}{0} \) specifically means that both the top and bottom of a fraction tend towards zero, making it unclear what the fraction should equal.
Common indeterminate forms include:

• \( \frac{0}{0} \)
• \( \frac{\infty}{\infty} \)
• \( 0 \times \infty \)
• \( 1^\infty \), among others.

To resolve these forms, mathematicians use tools like L'Hôpital's Rule or algebraic manipulation to "clear up" the indeterminacy and evaluate the limit.

Trigonometric Limits

Trigonometric limits involve special trigonometric functions like sine, cosine, and tangent. Often, the behavior of these functions as they approach certain values, such as 0, is critical in calculus.
When dealing with trigonometric limits, it's common to encounter the following scenarios:

• As \( t \to 0 \), \( \sin t \to 0 \) and \( \tan t \to 0 \).
• Meanwhile, \( \cos t \, \to 1 \).

In the given limit \( \lim _{t \to 0} \frac{2t}{\tan t} \), we're looking at a scenario where both the numerator and denominator approach zero. Therefore, this expression initially appears as an indeterminate form \( \frac{0}{0} \). By using L'Hôpital's Rule, we can differentiate the trigonometric components to assist with finding a definite limit.

Derivatives

Derivatives are a fundamental concept in calculus, measuring how a function changes as its input changes. Essentially, the derivative of a function at a point gives the slope of the tangent line to the function's graph at that point.
For basic functions:

• The derivative of \( f(t) = 2t \) is \( f'(t) = 2 \).
• The derivative of \( g(t) = \tan t \) is \( g'(t) = \sec^2 t \).

In applying L'Hôpital's Rule to resolve the indeterminate form \( \frac{0}{0} \), we differentiate the numerator and denominator functions separately, as shown above. These derivatives help us reframe and simplify the original limit, making it possible to evaluate. So, by substitution, we rewrite the limit as \( \lim _{t \to 0} \frac{2}{\sec^2 t} \), which simplifies further using the identity \( \sec t = \frac{1}{\cos t} \).
This demonstrates the power of derivatives in finding limits that initially seem elusive.