/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 Evaluate each limit (if it exist... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate each limit (if it exists). Use L'Hospital's rule (if appropriate). $$\lim _{x \rightarrow 0} \frac{\ln \sin x}{\ln \tan x}$$

Short Answer

Expert verified
The limit is 0.

Step by step solution

01

Identify the form

First, we need to identify the form of the limit as \( x \) approaches 0. Substitute 0 into the given expression: \( \frac{\ln \sin x}{\ln \tan x} \). Both \( \ln \sin x \) and \( \ln \tan x \) approach \(-\infty\) as \( x \to 0^+ \) (since \( \sin x \approx x \) and \( \tan x \approx x \) when \( x \to 0 \)). This indicates an \( \frac{-\infty}{-\infty} \) indeterminate form.
02

Apply L'Hospital's Rule

Since the expression is in the indeterminate form \( \frac{-\infty}{-\infty} \), we can apply L'Hospital's Rule. Differentiate the numerator and denominator separately: Numerator derivative: \( \frac{d}{dx}(\ln \sin x) = \cot x \)Denominator derivative: \( \frac{d}{dx}(\ln \tan x) = \csc^2 x \)
03

Evaluate the limit after differentiation

Substitute these derivatives into the original limit:\[\lim _{x \rightarrow 0} \frac{\cot x}{\csc^2 x}\]This simplifies to:\[\lim _{x \rightarrow 0} \frac{\cos x / \sin x}{1/\sin^2 x} = \lim _{x \rightarrow 0} \cos x \cdot \sin x = \lim_{x \rightarrow 0} \sin x \cdot \cos x\]
04

Evaluate the simplified limit

Since both \( \sin x \) and \( \cos x \) are continuous functions, substitute \( x = 0 \):\[\lim_{x \rightarrow 0} \sin x \cdot \cos x = \sin(0) \cdot \cos(0) = 0 \cdot 1 = 0\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limit Evaluation
When finding the limit evaluation of a function, especially as it approaches a specific value, we aim to understand the behavior of the function near that value. The goal is to determine what the output of the function's expression tends towards. For limits involving more complex functions, like natural logarithms or trigonometric functions, evaluating the end behavior accurately can get tricky. In such cases, L'Hospital's Rule becomes a handy tool to simplify and evaluate these limits. We apply this rule when the expression results in an indeterminate form, providing a structured means to handle seemingly undefined expressions as they approach certain values. Limit evaluation is crucial in calculus as it provides foundational understanding for continuity and the derivative concept.
Indeterminate Forms
Indeterminate forms occur when substituting values directly into a limit expression leads to undefined or ambiguous results, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). In the exercise, substituting \( x = 0 \) into \( \frac{\ln \sin x}{\ln \tan x} \) results in the form \( \frac{-\infty}{-\infty} \), which signals an indeterminate form. This form indicates a deeper evaluation is required to determine the limit's true behavior as \( x \) approaches the specified value. Differentiation via L'Hospital's Rule can be applied to handle these indeterminate forms, offering a clearer pathway to simplify and solve the limit expression. Without recognizing and addressing indeterminate forms, finding accurate limits would be challenging, hindering the progress in analyzing functions effectively.
Differentiation
Differentiation is the process of finding a derivative, indicating how a function changes as its input changes. In the context of L'Hospital's Rule, differentiation replaces indeterminate forms with a more manageable expression. Here, the expressions \( \ln \sin x \) and \( \ln \tan x \) are differentiated, resulting in \( \cot x \) and \( \csc^2 x \) respectively. These derivatives reflect the rates of change of the logarithmic functions involved in the original expression. Differentiation is crucial in this context as it transforms a complex limit into a simpler form that can be directly evaluated. The magic of calculus, specifically through differentiation, gives us the tools to crack indeterminate limits that are initially too complex to solve by mere substitution.
Trigonometric Functions
Trigonometric functions such as \( \sin x \), \( \cos x \), and \( \tan x \) are pivotal in calculus, especially when evaluating limits. These functions often exhibit predictable behavior as their arguments approach specific values. For \( x \to 0 \), the approximations \( \sin x \approx x \) and \( \tan x \approx x \) simplify complex expressions like the ones in our exercise. Within the differentiation process, these approximations help to untangle the multi-layered expressions. The derivatives involving trigonometric identities, namely \( \cot x = \frac{\cos x}{\sin x} \) and \( \csc^2 x = \frac{1}{\sin^2 x} \), allow us to reach an expression that simplifies beautifully. Understanding and using these trigonometric relationships are key in unraveling complex limits which incorporate trigonometric components.

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Most popular questions from this chapter

Use the following information. The hyperbolic sine of \(u\) is defined as \(\sinh u=\frac{1}{2}\left(e^{u}-e^{-u}\right)\) Figure 27.37 shows the graph of \(y=\sinh x\) The hyperbolic cosine of \(u\) is defined as \(\cosh u=\frac{1}{2}\left(e^{\mu}+\mathrm{e}^{-u}\right)\) Figure 27.38 shows the graph of \(y=\cosh x\) These functions are called hyperbolic functions since, if \(x=\cosh u\) and \(y=\sinh u, x\) and \(y\) satisfy the equation of the hyperbola \(x^{2}-y^{2}=1\) Verify the fact that the exponential expressions for the hyperbolic sine and hyperbolic cosine given above satisfy the equation of the hyperbola.

Find the derivatives of the given functions. $$y=4 x \sin 3 x$$

Use the following information. The hyperbolic sine of \(u\) is defined as \(\sinh u=\frac{1}{2}\left(e^{u}-e^{-u}\right)\) Figure 27.37 shows the graph of \(y=\sinh x\) The hyperbolic cosine of \(u\) is defined as \(\cosh u=\frac{1}{2}\left(e^{\mu}+\mathrm{e}^{-u}\right)\) Figure 27.38 shows the graph of \(y=\cosh x\) These functions are called hyperbolic functions since, if \(x=\cosh u\) and \(y=\sinh u, x\) and \(y\) satisfy the equation of the hyperbola \(x^{2}-y^{2}=1\) Show that \(\frac{d}{d x} \sinh u=\cosh u \frac{d u}{d x}\) and \(\frac{d}{d x} \cosh u=\sinh u \frac{d u}{d x}\) where \(u\) is a function of \(x\)

Find the differentials of the given functions. $$y=2.5 \sec ^{3} 2 t$$

Solve the given problems. The number \(N\) of reflections of a light ray passing through an optic fiber of length \(L\) and diameter \(d\) is \(N=\frac{L \sin \theta}{d \sqrt{n^{2}-\sin ^{2} \theta}}\) Here, \(n\) is the index of refraction of the fiber, and \(\theta\) is the angle between the light ray and the fiber's axis. Find \(d N / d \theta\)

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