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Evaluate the limit using l'Hôpital's Rule if appropriate. $$ \lim _{x \rightarrow 0^{+}} \frac{\ln x}{2+3 \ln (\sin x)} $$

Short Answer

Expert verified
The short answer to the problem is: $$ \lim_{x \rightarrow 0^{+}} \frac{\ln x}{2+3 \ln (\sin x)} = \frac{1}{3} $$

Step by step solution

01

Check l'Hôpital's Rule conditions

Before applying l'Hôpital's rule, we need to verify that the limit is in an indeterminate form.
02

Step 1.1: Evaluate the numerator

As \(x\) approaches 0 from the positive side, the natural logarithm of \(x\) tends to \(-\infty\): $$ \lim_{x \rightarrow 0^{+}} \ln x = -\infty $$
03

Step 1.2: Evaluate the denominator

As \(x\) approaches 0 from the positive side, the sine function approaches 0, and the natural logarithm of the sine function also tends to \(-\infty\): $$ \lim_{x \rightarrow 0^{+}} \ln (\sin x) = -\infty $$ This gives \(\lim_{x \rightarrow 0^{+}} (2+3 \ln (\sin x)) = -\infty\), since 2 is a constant and doesn't affect the limit.
04

Step 1.3: Determining the limit form

Our limit is in the \(\frac{(-\infty)}{(-\infty)}\) form: $$ \lim_{x \rightarrow 0^{+}} \frac{\ln x}{2+3 \ln (\sin x)} = \frac{-\infty}{-\infty} $$ Since our limit is in the indeterminate form and our functions are differentiable, we can apply l'Hôpital's rule.
05

Apply l'Hôpital's Rule

Applying l'Hôpital's rule, we differentiate both the numerator and denominator separately with respect to \(x\): $$ \lim_{x \rightarrow 0^{+}} \frac{\ln x}{2+3 \ln (\sin x)} = \lim_{x \rightarrow 0^{+}} \frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(2+3 \ln (\sin x))} $$
06

Differentiate both numerator and denominator

Differentiating the natural logarithm of \(x\): $$ \frac{d}{dx}(\ln x) = \frac{1}{x} $$ Now, differentiate the denominator using the chain rule: $$ \frac{d}{dx}(2+3 \ln (\sin x)) = \frac{d}{dx}(3 \ln (\sin x)) = 3\frac{d}{dx}(\ln (\sin x)) = 3\frac{cos(x)}{\sin x} $$
07

Substitute the derivatives into the limit expression

Substitute the derivatives back into the limit expression: $$ \lim_{x \rightarrow 0^{+}} \frac{\ln x}{2+3 \ln (\sin x)} = \lim_{x \rightarrow 0^{+}} \frac{\frac{1}{x}}{3\frac{cos(x)}{\sin x}} $$
08

Simplify the expression and evaluate the limit

First, simplify the expression: $$ \lim_{x \rightarrow 0^{+}} \frac{\frac{1}{x}}{3\frac{cos(x)}{\sin x}} = \lim_{x \rightarrow 0^{+}} \frac{\sin x}{3x \cos x} $$ Now, as \(x \rightarrow 0^{+}\), we have: $$ \lim_{x \rightarrow 0^{+}} \frac{\sin x}{3x \cos x} = \frac{\lim_{x \rightarrow 0^{+}} \frac{\sin x}{x}}{\lim_{x \rightarrow 0^{+}} 3\cos x} $$ Using the standard limit result \(\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1\) and evaluating the cosine limit directly, we get: $$ \lim_{x \rightarrow 0^{+}} \frac{\sin x}{3x \cos x} = \frac{1}{3} $$ So, the limit of the given expression is \(\frac{1}{3}\).

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Key Concepts

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

Limits
Limits are a fundamental concept in calculus, which involve finding the value that a function approaches as the input tends towards a particular point. In this exercise, we are looking at the limit as \( x \) tends to \( 0^{+} \), which means from the positive side. This kind of one-sided limit helps us understand behavior that might differ if approached from negative direction.
When calculating limits, you might encounter specific forms that suggest indeterminacy, where typical evaluation methods fail. That's where special rules like l'Hôpital's Rule can come in handy. The idea is to transform the problem into one that's easier to handle, by differentiating the numerator and denominator to simplify these indeterminate forms.
Evaluating limits correctly often involves checking these conditions and using proper techniques, such as substitution, multiplication, or differentiation, to find a coherent, finite answer.
Indeterminate Forms
Indeterminate forms arise in limits when straightforward substitution yields expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms do not lead directly to any clear answer. In the exercise provided, the limit initially takes the form of \( \frac{-\infty}{-\infty} \).
These scenarios require additional techniques to evaluate properly. Using l'Hôpital's Rule can resolve these forms by replacing them with derivatives that are easier to evaluate. This rule applies specifically to fractions resulting in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) forms.
By differentiating the top and bottom separately, you can often convert the form into something solvable. It's a powerful tool in calculus for navigating these tricky expressions. Be aware: confirming the conditions for indeterminate forms first is critical before applying such methods.
Differentiation
Differentiation is a core operation in calculus, allowing us to compute the rate of change of a function. In this exercise, differentiation is used both in the numerator and the denominator during the application of l'Hôpital's Rule.
When differentiating the natural logarithm of \( x \), the derivative is quite simple: \( \frac{1}{x} \). On the other hand, the chain rule must be applied to differentiate \( 3 \ln(\sin x) \). The derivative here is \( 3 \frac{\cos x}{\sin x} \), where \( \frac{d}{dx} \ln(\sin x) \) becomes \( \frac{\cos x}{\sin x} \). Using such rules correctly is essential for finding accurate derivatives.
Differentiation transforms the original problem into an equivalent form, whose limits are often simpler to evaluate. This makes it a vital technique in calculus, providing insights into the behavior of complex functions.
Natural Logarithm
The natural logarithm is a logarithm to the base \( e \), an irrational and transcendental number approximately equal to 2.718. It is commonly used in various fields of science and mathematics due to its natural growth properties.
In differentiation, the natural logarithm function \( \ln x \) has a simple derivative, \( \frac{1}{x} \). In this exercise, both the natural logarithm of \( x \) and \( \sin x \) appear. Each impacts the function's rate of change as \( x \) approaches a limit.
Understanding the math behind the natural logarithm's behavior near limits helps solve this and similar exercises. Recognizing the logarithmic form can simplify complex expressions, as it inherently incorporates exponential relationships that are instrumental in calculus analyses.

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