Problem 39 - Calculus - AP Edition

Question

Find the limits. $$\lim _{x \rightarrow 0^{+}} x^{\sin x}$$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Rewrite the Expression

The expression given is \( x^{\sin x} \). To find the limit, let's take the natural logarithm of the expression so we can bring the exponent down as a multiplier. This gives us \( \ln(y) = \sin x \cdot \ln x \) where \( y = x^{\sin x} \). We need to evaluate \( \lim_{x \to 0^+} \ln(y) \).

Step 2: Evaluate the New Limit

Focus on the expression \( \sin x \cdot \ln x \). As \( x \to 0^+ \), \( \sin x \to 0 \), and \( \ln x \to -\infty \). Analyze \( \lim_{x \to 0^+} \sin x \cdot \ln x \), which is an indeterminate form of type \( 0 \times -\infty \). Rewrite it to be suitable for L'Hôpital's Rule.

Step 3: Apply L'Hôpital's Rule

Rewrite \( \sin x \cdot \ln x \) as \( \frac{\ln x}{1/\sin x} \) to create the \( \frac{-\infty}{\infty} \) indeterminate form. Apply L'Hôpital's Rule:Find the derivatives:- Numerator: \( \frac{d}{dx}(\ln x) = \frac{1}{x} \)- Denominator: \( \frac{d}{dx}(1/\sin x) = -\frac{\cos x}{\sin^2 x} \)So, apply L'Hôpital's Rule:\[ \lim_{x \to 0^+} \frac{\ln x}{1/\sin x} = \lim_{x \to 0^+} \frac{1/x}{-\cos x/\sin^2 x} = \lim_{x \to 0^+} -\frac{\sin^2 x}{x \cos x} \]

Step 4: Simplify the Expression

The limit now simplifies to \( -\frac{\sin^2 x}{x \cos x} \). Note that \( \lim_{x \to 0^+} \frac{\sin^2 x}{x} = \lim_{x \to 0^+} \frac{\sin x \cdot \sin x}{x} \). Utilize \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). Thus:\[ \lim_{x \to 0^+} \frac{\sin x}{x} \rightarrow 1 \]So the limit can be simplified using \( \cos x \to 1 \):\[ \lim_{x \to 0^+} -\frac{\sin^2 x}{x \cos x} = 0 \]

Step 5: Revert the Logarithm

The limit \( \lim_{x \to 0^+} \ln y = 0 \) implies that \( y = x^{\sin x} \) goes to \( e^0 = 1 \) as \( x \) approaches \( 0^+ \). Hence, the original limit is:\[ \lim_{x \to 0^+} x^{\sin x} = 1 \]

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a valuable tool for evaluating limits involving indeterminate forms. When you encounter a limit that results in an undefined expression like \( \frac{0}{0} \) or \( \frac{-\infty}{\infty} \), L'Hôpital's Rule can help you find the limit more easily.

According to the rule, if \( \lim_{x \to c} \frac{f(x)}{g(x)} \) results in such an indeterminate form, then you can take the derivatives of the numerator and the denominator separately to find the limit:

• Compute \( f'(x) \) and \( g'(x) \).
• Re-evaluate the limit as \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \).

In the context of this exercise, L'Hôpital's Rule transforms \( \frac{\ln x}{1/\sin x} \) into a more manageable expression to evaluate the limit.

Indeterminate Forms

Indeterminate forms arise when calculating limits leads to undefined mathematical expressions. Common types include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), and \( 0 \times \infty \). These forms indicate that a straightforward calculation of limits isn't possible.

In this exercise, the expression \( \sin x \cdot \ln x \) approaches an indeterminate form when \( x \to 0^+ \). Specifically, you encounter \( 0 \times -\infty \), which makes it necessary to rewrite the expression using algebraic manipulation. This gives you the opportunity to use L'Hôpital's Rule effectively, converting the expression into a quotient for differentiation.

Natural Logarithm

The natural logarithm, denoted \( \ln \), is a logarithm to the base \( e \), where \( e \approx 2.718 \). It is a critical function in calculus, particularly for simplifying complex expressions involving exponents. One key property of the natural logarithm is \( \ln(e^x) = x \), illustrating its inverse relationship with the exponential function.

In this problem, taking the natural logarithm of \( x^{\sin x} \) allows for the simplification of the exponent, transforming \( x^{\sin x} \) into a product \( \sin x \cdot \ln x \). This transformation is crucial for evaluating the limit and using tools like L'Hôpital's Rule. The natural logarithm essentially brings the exponent down to a level where algebraic manipulation becomes possible.

Trigonometric Limits

Trigonometric limits often involve expressions where trigonometric functions like sine or cosine appear near points of discontinuity or indeterminate forms, such as \( x \to 0 \). A classical example is the limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), which is foundational in calculus.

In this exercise, analyzing \( \sin x \cdot \ln x \) as \( x \to 0^+ \) involves understanding the behavior of \( \sin x \). This function approaches zero in a predictable pattern around zero. In the last step of the solution, the fact that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) is utilized to simplify the limit expression \( -\frac{\sin^2 x}{x \cos x} \) and find that the result of the original limit is 1. By acknowledging these foundational limits, solving more complex problems becomes more manageable.