Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each limit. $$ \lim _{t \rightarrow 0} \frac{\tan 2 t}{\sin 2 t-1} $$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Understanding the Problem

We need to evaluate the limit \( \lim _{t \rightarrow 0} \frac{\tan 2 t}{\sin 2 t-1} \). At \( t=0 \), we'll encounter indeterminate form \( \frac{0}{0} \), requiring L'Hôpital's Rule or algebraic manipulation.

Step 2: Initial Substitution

Substitute \( t = 0 \) into the expression to confirm the indeterminate form. We have \[ \tan(2 \cdot 0) = 0, \quad \sin(2 \cdot 0) - 1 = -1, \] thus the limit form looks like \( \frac{0}{-1} \), clearly manifesting indeterminacy as \( t \to 0 \).

Step 3: Apply L'Hôpital's Rule

Since the limit yields an indeterminate form, apply L'Hôpital's Rule. Take the derivatives of the numerator and the denominator:- Derivative of numerator \( d/dt[\tan(2t) ] = 2\sec^2(2t) \).- Derivative of denominator \( d/dt[\sin(2t) - 1] = 2\cos(2t) \).

Step 4: Calculate New Limit

Evaluate the new limit after substitution:\[ \lim_{t \to 0} \frac{2\sec^2(2t)}{2\cos(2t)} = \lim_{t \to 0} \frac{\sec^2(2t)}{\cos(2t)}. \]

Step 5: Simplify and Evaluate

Simplify \( \frac{\sec^2(2t)}{\cos(2t)} = \frac{1}{\cos(2t)} \cdot \frac{1}{\cos(2t)} = \frac{1}{\cos^3(2t)} \). Since \( \cos(2 \cdot 0) = 1 \), the expression becomes \( \frac{1}{1^3} = 1 \). So the limit is 1.

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is an essential tool in calculus for dealing with limits that lead to indeterminate forms. It simplifies the process of finding limits for functions that result in expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When encountering these forms, you can apply the rule by

• differentiating the numerator and
• then the denominator separately.

After taking the derivatives, you evaluate the limit again to see if it resolves the indeterminate form. If not, you may need to apply the rule multiple times. It's crucial that both the original limits \( \lim_{x \to c} f(x)\) and \(\lim_{x \to c} g(x)\) yield the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) before using the rule. This technique simplifies complex problems, making it easier to find the limit.

Indeterminate Forms

Indeterminate forms arise when substituting a value into a function results in an undefined expression. In calculus, expressions like \( \frac{0}{0} \), \( 0 \times \infty \), or \( \infty - \infty \) count as indeterminate, requiring special methods to address them. When dealing with limits, if direct substitution leads to an indeterminate form, you cannot conclusively determine the limit without further manipulation.

• One common method is L'Hôpital's Rule, where we differentiate the numerator and the denominator independently to resolve these forms.
• Another approach may involve algebraic simplification, such as factoring or expanding expressions, to avoid the indeterminate form.

Understanding indeterminate forms is fundamental, as they indicate that the typical route of evaluating limits (like simple substitution) will not provide a direct answer.

Trigonometric Limits

Trigonometric limits often involve functions like sine, cosine, tangent, and their derivatives, making them a critical part of calculus. Such limits can initially seem complex due to the periodic and oscillating nature of these functions.

• For small values of \( t \), expansions or identities, such as \( \lim_{t \to 0} \frac{\sin(t)}{t} = 1 \), become vital tools for simplification.
• Special limits where trigonometric functions are evaluated as \( t \to 0 \) or other significant points often employ these identities and rules.

When solving trigonometric limits, it's useful to be familiar with key identities and derivatives—such as the derivative of tangent, \( \frac{d}{dt}[\tan(t)] = \sec^2(t) \). Grasping these concepts allows for smoother calculation of limits involving trigonometric functions, paving the way for understanding broader calculus topics.