Problem 70 - Thomas Calculus in SI Units

Question

L'Hópital's Rule does not help with the limits. Try it-you just keep on cycling. Find the limits some other way. $$\lim _{x \rightarrow 0^{+}} \frac{\cot x}{\csc x}$$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Simplify the Trigonometric Expression

Start by expressing the cotangent and cosecant functions in terms of sine and cosine. Since \( \cot x = \frac{\cos x}{\sin x} \) and \( \csc x = \frac{1}{\sin x} \), the expression becomes: \[\frac{\cot x}{\csc x} = \frac{ \frac{\cos x}{\sin x} }{ \frac{1}{\sin x} } = \cos x \cdot \sin x.\]

Step 2: Evaluate the Product at the Limit

Now evaluate the limit of the new expression \( \cos x \cdot \sin x \) as \( x \to 0^+ \). The cosine function is continuous, and at \( x = 0 \), \( \cos(0) = 1 \). Also, when \( x \to 0^+ \), \( \sin x \to 0 \). Thus the expression becomes:\[\lim_{x \to 0^+} \cos x \cdot \sin x = \lim_{x \to 0^+} (1 \cdot 0) = 0.\]

Step 3: Conclusion of the Limit

Given that both trigonometric functions are continuous around \( x = 0 \), and we have calculated \( \cos x \to 1 \) and \( \sin x \to 0 \), the limit of their product as \( x \to 0^+ \) is 0. Thus, we find without needing to repeatedly apply L'Hôpital's Rule: \[\lim_{x \to 0^+} \frac{\cot x}{\csc x} = 0.\]

Key Concepts

Trigonometric Limits

Understanding trigonometric limits is a crucial part of mastering calculus. Trigonometric functions often appear in limit problems, especially involving scenarios where direct substitution would lead to an indeterminate form. To handle such problems, it's essential to rewrite these functions using their basic identities.

In the given problem, we deal with cotangent and cosecant functions. These trigonometric functions can be expressed in terms of sine and cosine because:

• The cotangent function is defined as \( \cot x = \frac{\cos x}{\sin x} \).
• The cosecant function is defined as \( \csc x = \frac{1}{\sin x} \).

These identities allow us to simplify complex expressions and make it easier to evaluate limits. By rewriting the given limit expression involving \( \frac{\cot x}{\csc x} \) using these identities, it can be simplified to \( \cos x \cdot \sin x \), directly enabling us to evaluate the limit as \( x \to 0^+ \). This simplification is a powerful example of how knowing your trigonometric identities can simplify otherwise complex calculus problems. Always remember, simplifying the expression is often your best first step when dealing with trigonometric limits.

Limit Evaluation

When evaluating limits, especially in trigonometry, substituting the variable directly into the expression can often yield valuable insights. However, it sometimes leads to expressions like \( \frac{0}{0} \), which are indeterminate forms.

In evaluating the limit \(\lim_{x \to 0^+} \frac{\cot x}{\csc x}\), simplification led to \( \cos x \cdot \sin x \). Now, as \( x \to 0^+ \):

• \( \cos x \) approaches \( \cos(0) = 1 \).
• \( \sin x \) approaches \( 0 \).

Evaluating the limit, \( \lim_{x \to 0^+} \cos x \cdot \sin x = \lim_{x \to 0^+} (1 \cdot 0) = 0 \). The approach demonstrates not only the simplification but also how continuity of trigonometric functions aids in evaluating limits. Using direct substitution in trigonometric limits, especially after simplification, commonly resolves the need for more complex calculus techniques like L'Hôpital's Rule. When faced with trigonometric limits, always consider simple substitution after you simplify the expression.

Calculus Techniques

Calculus techniques provide various tools for tackling limits. One popular method is L'Hôpital's Rule, used when you encounter indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It involves taking the derivative of the numerator and the denominator separately, then evaluating the limit again.

In the exercise above, however, L'Hôpital's Rule doesn't help because the expressions cycle through the same problem after differentiation. Instead, the simplification of the expression to \( \cos x \cdot \sin x \) offers a straightforward path to evaluating the limit.

• Simplification can instantly resolve the indeterminacy by converting complex trigonometric functions into more manageable forms.
• Substituting the variable within continuous trigonometric functions often yields the limit without further computation.
• Understanding basic trigonometric identities and their behavior near specific points can simplify calculus problems tremendously.

When direct evaluation or basic identity use doesn't suffice, calculus techniques like L'Hôpital's Rule come into play. But always first look to simplify the expression, as in this case, it provided the correct result without the need for more elaborate methodologies. This highlights the importance of choosing the right technique for the problem at hand.