Question

Find each limit. Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$ \lim _{x \rightarrow 0^{+}} \frac{\ln \sin ^{2} x}{3 \ln \tan x} $$

Short answer

The limit is 0.

Step-by-step solution

Identify the Form

First, consider the expression \( \ln \sin^2 x \) and \( 3 \ln \tan x \) as \( x \to 0^+ \). We know \( \sin x \to 0 \) and \( \tan x \to 0 \), hence \( \ln \sin^2 x \to -\infty \) and \( 3 \ln \tan x \to -\infty \). Therefore, the expression has the indeterminate form \( \frac{-\infty}{-\infty} \), permitting the use of l'Hôpital's Rule.

Apply l'Hôpital's Rule

Apply l'Hôpital's Rule, which requires us to differentiate the numerator and the denominator separately. The derivative of the numerator \( \ln \sin^2 x \) is \( \frac{2 \cos x}{\sin x} \) or equivalently \( 2 \cot x \). The derivative of the denominator \( 3 \ln \tan x \) is \( 3 \cdot \frac{1}{\tan x} \cdot \sec^2 x \), which simplifies to \( 3 \cdot \frac{1}{\sin x \cdot \cos x} \).

Simplify and Evaluate the Limit

After differentiation, the limit becomes \[ \lim _{x \rightarrow 0^{+}} \frac{2 \cot x}{3 \cdot \frac{1}{\sin x \cdot \cos x}}. \] Simplifying this gives \[ \lim _{x \rightarrow 0^{+}} \frac{2 \cos x \cdot \cos x \cdot \sin x}{3} = \lim _{x \rightarrow 0^{+}} \frac{2 \cos^2 x \cdot \sin x}{3}. \] As \( x \to 0^+ \), \( \cos^2 x \to 1 \) and \( \sin x \to 0 \). Hence, the whole expression simplifies to \( \frac{0}{3} = 0 \).

Key concepts

Indeterminate Forms

Indeterminate forms are expressions that arise in calculus when an operation, such as division or exponentiation, gives an ambiguous result. Several types of indeterminate forms exist, including \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( \infty - \infty \), \( 0 \times \infty \), \( 1^\infty \), \( 0^0 \), and \( \infty^0 \). These forms, especially in limits, do not directly indicate the outcome and require further analysis.

In the given exercise, the expression \( \frac{\ln \sin^2 x}{3 \ln \tan x} \) as \( x \to 0^+ \) is a classic case of the indeterminate form \( \frac{-\infty}{-\infty} \). Both the numerator and denominator trend towards \(-\infty\). This situation suggests ambiguity because knowing that both tend to \(-\infty\) does not provide a clear indication of the limit value.

By using techniques like L'Hôpital's Rule, one can resolve this indeterminacy by differentiating the numerator and denominator separately.

Calculus Limits

Calculus limits are foundational in understanding the behavior of functions as they approach specific points or infinities. Limits help in defining concepts like continuity, derivatives, and integrals.

To solve limits involving indeterminate forms, such as the above exercise, one often applies calculus techniques to discover more manageable expressions. The process of taking limits often involves substitution, manipulation, or special rules like L'Hôpital's Rule.

For the expression \( \lim_{x \to 0^+} \frac{\ln \sin^2 x}{3 \ln \tan x} \), as \( x \) approaches zero from the positive side, both the natural logarithms become negatively infinite, leading us to use calculus tools to simplify. Intuition alone won't suffice, hence differentiation permits refinement of the expression, turning complex behavior into plain numerical results.

Differentiation

Differentiation is a method used in calculus to determine the rate at which a function is changing at any given point. It is the process of finding a derivative.

• The derivative of a function gives the slope of the tangent line at any point on the curve.
• Used widely in finding limits, differentiation transforms complex expressions into simpler forms.

In this problem, we differentiate \( \ln \sin^2 x \) and \( 3 \ln \tan x \) to apply L'Hôpital's Rule.

The derivative of \( \ln \sin^2 x \) involves using the chain rule, resulting in \( 2 \cot x \). For the denominator, \( 3 \ln \tan x \), we apply differentiation rules to obtain \( 3 \cdot \frac{1}{\sin x \cdot \cos x} \). This process changes the complexity from negative infinity evaluations to functions involving trigonometric identities, paving the way for easy limit calculation.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, are crucial in the study of calculus. These functions describe angles and relationships in triangles but extend to model periodic phenomena applicable throughout mathematics and physics.

• \(\sin x\) and \(\tan x\) are key in this exercise, where their behavior near zero impacts the indeterminate form.
• The identities and relationships between these functions assist in simplifying expressions.

In our limit problem, as \(x\) approaches \(0^+\), \(\sin x\) and \(\tan x\) approach zero differently, which drives the \(-\infty\) outcomes for logarithms. Being adept with identities like \(\cot x = \frac{1}{\tan x}\) and knowing trigonometric limits, such as \(\sin x \approx x\) near zero, help in transitioning from raw expressions to evaluated limits.

Understanding these functions' derivatives allows applying L'Hôpital's Rule effectively, navigating through intricacies to achieve a simplified, precise limit value.