Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$\lim _{x \rightarrow 0}\left(\csc ^{2} x-\cot ^{2} x\right)$$

Step-by-Step Solution

Answer

1

Step 1: Expression in terms of sine and cosine

To apply l'Hôpital's Rule, we first need to express \( \csc^2 x \) and \( \cot^2 x \) in terms of sine and cosine. \( \csc x \) is the reciprocal of \( \sin x \), so \( \csc^2 x = \frac{1}{\sin^2 x} \). Likewise, \( \cot x = \frac{\cos x}{\sin x} \), so \( \cot^2 x = \frac{\cos^2 x}{\sin^2 x} \). Thus, the expression becomes \[ \lim _{x \rightarrow 0}\left( \frac{1}{\sin^2 x} - \frac{\cos^2 x}{\sin^2 x} \right). \]

Step 2: Simplify the expression

Combine the fractions: \[ \lim _{x \rightarrow 0} \frac{1 - \cos^2 x}{\sin^2 x}. \] Note that \( 1 - \cos^2 x = \sin^2 x \) (from the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \)). This simplifies the expression to \[ \lim _{x \rightarrow 0} \frac{\sin^2 x}{\sin^2 x} = \lim _{x \rightarrow 0} 1. \]

Step 3: Evaluate the limit directly

The expression has simplified to \( \lim _{x \rightarrow 0} 1 \), which is clearly 1 since this limit does not depend on \( x \). Thus, the final answer is \[ \boxed{1}. \]

Key Concepts

Indeterminate form

An indeterminate form is a mathematical expression that doesn't immediately reveal its limit. When evaluating limits, it’s crucial to identify whether an expression stands as indeterminate. Common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), and \( 0 \cdot \infty \). These forms necessitate further manipulation or techniques, such as l'Hôpital's Rule, to resolve.
When approaching this particular limit, \( \lim_{x \rightarrow 0}\left( \csc^2 x - \cot^2 x\right) \), initial substitution shows an indeterminate form as both \( \csc^2 x \) and \( \cot^2 x \) approach infinity as \( x \) tends to zero. Therefore, expressing them in terms of sine and cosine is a crucial first step to simplify the problem and determine the true behavior of the expression around zero.

Trigonometric identities

Trigonometric identities are relationships between trigonometric functions that hold true for all values of the involved variables. These identities are powerful tools in simplifying expressions and solving equations.
In relation to the indeterminate form \( \lim _{x \rightarrow 0}\left( \csc ^{2} x - \cot ^{2} x\right) \), we utilize the following trigonometric identities:

• \( \csc x = \frac{1}{\sin x} \)
• \( \cot x = \frac{\cos x}{\sin x} \)
• Pythagorean identity: \( \sin^2 x + \cos^2 x = 1 \)

Replacing \( \csc x \) and \( \cot x \) with their sine and cosine equivalents reveals a path to simplify the expression. Combining these into a single fraction helps us recognize a known identity: \( 1 - \cos^2 x = \sin^2 x \). This step moves us past the indeterminate form toward a solvable limit.

Limit evaluation

Limit evaluation is the process of determining the value that a function approaches as the input (or 'x') approaches some value. Limits are essential in calculus for understanding continuity, derivatives, and integrals. When we check \( \lim_{x \rightarrow 0}\left( \csc^2 x - \cot^2 x\right) \), our initial simplification through trigonometric identities lands us with \( \lim _{x \rightarrow 0} \frac{\sin^2 x}{\sin^2 x} \).
Recognizing that we have \( \frac{\sin^2 x}{\sin^2 x} = 1 \), we can now directly evaluate this limit. Since the expression simplifies to a constant, further calculations are unnecessary. Thus, \( \lim_{x \rightarrow 0} 1 \) becomes clear and provides us with the straightforward result of \( 1 \), showing how simplification and proper limit rules swiftly resolve the original indeterminate expression.