Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$\lim _{x \rightarrow 0^{+}}(\sin x)^{x}$$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify the Limiting Form

First, substitute the approach value in the expression to determine its form. Substituting 0 into the limit, we have \( \lim_{x \to 0^+} (\sin x)^{x} = (\sin 0^+)^{0} = 0^0 \). This is considered an indeterminate form.

Step 2: Transform for l'Hôpital's Rule

Rewrite the expression into a form applicable for l'Hôpital's Rule. Let \( y = (\sin x)^{x} \). Take the natural logarithm on both sides to obtain \( \ln y = x \ln(\sin x) \). Now we have an expression that is \( \lim_{x \to 0^+} x \ln(\sin x) \). This limit is of type \( 0 \cdot (-\infty) \), so rewrite it as \( \lim_{x \to 0^+} \frac{\ln(\sin x)}{1/x} \).

Step 3: Apply l'Hôpital's Rule

With the expression transformed, apply l'Hôpital's Rule to \( \lim_{x \to 0^+} \frac{\ln(\sin x)}{1/x} \). We need to check if the form \( \frac{-\infty}{\infty} \) is applicable, which it is. Differentiate the numerator and the denominator independently and calculate the limit: \[ \lim_{x \to 0^+} \frac{(\sin x)'}{(\sin x)} \cdot \frac{x}{-x^{-2}} = \lim_{x \to 0^+} \frac{\cos x}{\sin x} \] which simplifies to \( \lim_{x \to 0^+} \cot x. \)

Step 4: Evaluate Trigonometric Limit

To calculate \( \lim_{x \to 0^+} \cot x \, \), observe that \( \cot x = \frac{1}{\tan x}\) where \( \tan x \to 0^+ \) as \( x \to 0^+ \). Hence, \( \cot x \to \infty \). Therefore, \( \ln y \to -\infty \).

Step 5: Convert Back from Logarithm

Since \( \ln y \to -\infty \), it means \( y = e^{\ln y} \rightarrow e^{- ext{large number}} = 0 \). Therefore, \( \lim_{x \to 0^+} (\sin x)^{x} = 0. \)

Key Concepts

Indeterminate Forms

Indeterminate forms often arise in calculus when evaluating limits. They occur when substituting the limiting value gives a form that does not readily yield a specific result, like 0/0, ∞/∞, or 0⁰. These forms indicate that further analysis is required to determine the true behavior of the limit.
In the exercise, when substituting 0 into \(\lim_{x \to 0^+} (\sin x)^x\), we initially find the form 0⁰. This is indeterminate and suggests that simply plugging in the values won't work without additional manipulation.
Understanding indeterminate forms is crucial for correctly applying rules like l'Hôpital's Rule, which can simplify and solve tricky limits.

Limits in Calculus

Limits are fundamental to calculus and help understand the behavior of functions as variables approach a certain value. They form the basis for defining derivatives and integrals.
When finding limits, particularly those resulting in indeterminate forms, we often need extra techniques like algebraic manipulation, rationalization, or rules such as l'Hôpital's Rule to solve them effectively.
In the provided solution, after recognizing the indeterminate form, the limit expression \(\lim_{x \to 0^+} (\sin x)^x\) was transformed, highlighting the importance of limit transformations in solving calculus problems.

Trigonometric Limits

Trigonometric limits involve functions like sine, cosine, tangent, etc., and their interrelations as variables change. These are crucial in many calculus problems where trigonometric identities help simplify expressions.
The exercise involves \(\lim_{x \to 0^+} \cot x\), a trigonometric limit. Recognizing that \(\cot x = \frac{1}{\tan x}\) and as \(x\to 0^+\), \(\tan x o 0^+\) means \(\cot x o \infty\), helps solve the problem.
Mastering trigonometric limits enhances one's ability to tackle more complex calculus challenges by leveraging these fundamental properties.

Natural Logarithm Transformation

Natural logarithm transformations are powerful tools for simplifying limits, especially when encountering exponential expressions. By taking the natural logarithm of both sides, expressions often become more manageable.
In our exercise, transforming \(y = (\sin x)^{x}\) using \(\ln y = x \ln(\sin x)\) allowed the application of calculus techniques to solve the limit, particularly because it converted a complicating exponent into a product.
Students should become comfortable with using logarithms to open up new pathways for evaluating limits, especially indeterminate forms.