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: Understand the Limit Expression

We need to find the limit as \( t \) approaches 0 for the expression \( \frac{\tan 2t}{\sin 2t - 1} \). First, substitute \( t = 0 \) into the expression. Both \( \tan(0) \) and \( \sin(0) \) are 0, leading the expression \( \frac{0}{0 - 1} = 0 \). Since the substitution results in division by zero, we need a different approach to evaluate the limit.

Step 2: Apply Trigonometric Identities

Utilize the trigonometric identity \( \tan(2t) = \frac{\sin(2t)}{\cos(2t)} \) to rewrite the expression as \( \frac{\sin(2t)}{\cos(2t)(\sin(2t) - 1)} \). This transformation prepares the expression for further simplification or potential use of L'Hôpital's Rule.

Step 3: Consider L'Hôpital's Rule

Since direct substitution in the limit \( \lim_{t \to 0} \frac{\sin(2t)}{\cos(2t)(\sin(2t) - 1)} \) results in the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that \[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \] if the original limit is \( \frac{0}{0} \) and both \( f'(x) \) and \( g'(x) \) exist in a neighborhood around \( c \).

Step 4: Differentiate the Numerator and the Denominator

Differentiate the numerator \( \sin(2t) \) to get \( 2\cos(2t) \). Differentiate the denominator \( \cos(2t)(\sin(2t) - 1) \) using the product rule: \( f(x)g(x) \rightarrow f'(x)g(x) + f(x)g'(x) \). Here, \( f(t) = \cos(2t) \) and \( g(t) = \sin(2t) - 1 \). Thus, the derivative is \[ -2\sin(2t)(\sin(2t) - 1) + \cos(2t) \cdot 2\cos(2t) \].

Step 5: Evaluate the Limit Again

Substitute \( t = 0 \) into the differentiated numerator and denominator: The numerator becomes \( 2 \cdot \cos(0) = 2 \). For the denominator, \(-2 \cdot \sin(0) \cdot -1 + 2 \cdot \cos^2(0) = 0 + 2\cdot1 = 2\). Thus, the limit evaluates to \( \frac{2}{2} = 1 \).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus used to evaluate limits that result in indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When faced with these forms, directly substituting the variable into the function doesn't provide a clear answer. Instead, you can differentiate the numerator and the denominator separately, then take the limit again. This simplifies the problem and often yields a determinate answer.
It's crucial to confirm that the functional form is indeed an indeterminate form where both original functions' derivatives exist around the point you are considering. If these conditions aren't met, L'Hôpital's Rule cannot be applied.

Trigonometric Identities

Trigonometric identities are essential tools for simplifying expressions involving trigonometric functions. In the given problem, the expression \( \tan(2t) \) was rewritten using the identity \( \tan(x) = \frac{\sin(x)}{\cos(x)} \). This identity allows a complex trigonometric expression to be transformed into a form that is potentially easier to differentiate or manipulate.
Using these identities can often change the problem from unsolvable to straightforward. Knowing identities such as the Pythagorean identities or angle sum formulas can greatly simplify limit evaluation problems like this one.

Limit Evaluation

Evaluating limits involves finding what a function approaches as the input approaches a certain value. For trigonometric limits, substituting variables often leads to helpful insights or, at times, indeterminate forms. Initial substitution of the limit point helps to evaluate whether the limit results in an obvious answer or requires further work.
In this particular exercise, upon substitution, \( t = 0 \) led to an indeterminate form, signaling that a deeper look with either simplification or L'Hôpital's Rule was necessary. Using limits properly ensures this calculus method evaluates how functions behave unexpectedly around critical points.

Indeterminate Form

An indeterminate form arises when direct substitution into a mathematical expression does not lead to a well-defined limit. Common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), 0\cdot\infty, among others. In such cases, these forms indicate incomplete information for a straightforward evaluation.
The importance of recognizing indeterminate forms is vital as it often hints the need for additional techniques, like L'Hôpital's Rule or algebraic simplification, to find the value of the limit. In this exercise, recognizing \( \frac{0}{0} \) set the stage for using L'Hôpital's Rule effectively.

Calculus Problem Solving

Solving calculus problems involves several strategic steps, from identifying expressions' structures to applying the correct mathematical rules. It begins with understanding the problem and substituting values to see what form the function takes.
For our given limit, initial substitution led to an indeterminate form, prompting the use of trigonometric identities to simplify the expression, and ultimately applying L'Hôpital's Rule to evaluate the undefined expressions. Each move was a calculated step to simplify and find the correct solution. The chain of reasoning from substitution, through application of identities and rules, reflects the layered approach required in advanced calculus problem solving.