Question

Find the indicated limit. Make sure that you have an indeterminate form before you apply l'Hôpital's Rule. $$ \lim _{x \rightarrow 0} \frac{\int_{0}^{x} \sqrt{1+\sin t} d t}{x} $$

Short answer

The limit is 1.

Step-by-step solution

Finding the Limit Expression

We are given the expression to evaluate: \( \lim _{x \rightarrow 0} \frac{\int_{0}^{x} \sqrt{1+\sin t} \, dt}{x} \). Since both the numerator and denominator approach zero as \( x \rightarrow 0 \), this is an indeterminate form \( \frac{0}{0} \). This allows us to use l'Hôpital's Rule.

Applying l'Hôpital's Rule

To apply l'Hôpital's Rule, we need the derivatives of the numerator and denominator:1. The derivative of the numerator \( \int_{0}^{x} \sqrt{1+\sin t} \, dt \) is \( \sqrt{1+\sin x} \), by the Fundamental Theorem of Calculus.2. The derivative of the denominator \( x \) is \( 1 \).Thus, we transform the limit to \( \lim_{x \rightarrow 0} \frac{\sqrt{1+\sin x}}{1} \).

Evaluating the New Limit

Now we evaluate the transformed limit: \( \lim_{x \rightarrow 0} \sqrt{1+\sin x} \). As \( x \rightarrow 0 \), \( \sin x \rightarrow 0 \). Therefore, \( \sqrt{1+\sin x} \rightarrow \sqrt{1} = 1 \).

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a crucial concept that bridges the world of differentiation and integration. It's a two-part theorem that links these ideas in a seamless way.

Generally, this theorem states that if a function is integrable over an interval, then its integral function is differentiable and the derivative is the original function. In simple terms, if you accumulate the area under the curve of a function \( f(t) \) from a starting point up to a variable point \( x \), then the derivative of this accumulation with respect to \( x \) is simply \( f(x) \).

• First Part: The first component of the theorem guarantees that an antiderivative (or indefinite integral) of a function exists if it's continuous.
• Second Part: The second part allows us to compute definite integrals in terms of antiderivatives – which means if you have calculated an antiderivative, you can find the value of the integral by simply evaluating at two endpoints.

In practice, this theorem is extremely powerful as it enables us to evaluate integrals quickly using antiderivatives. In the given exercise, we use this to find the derivative of an integral expression easily, setting the stage to apply additional calculus techniques.

Indeterminate Forms

Indeterminate forms arise when calculating limits and the expression doesn't plainly resolve to a number. There are seven classic forms of indeterminacy in calculus, such as \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), \( \infty - \infty \), and others.

In the given exercise, as \( x \to 0 \), both the numerator, which is an integral expression, and the denominator \( x \) approach zero. This creates an indeterminate form \( \frac{0}{0} \), allowing us to utilize l'Hôpital's Rule for resolution.

• Using l'Hôpital's Rule helps us tackle these expressions by differentiating both the numerator and the denominator to see if the limit becomes more straightforward.
• It's important to clearly identify indeterminate forms before applying rules like l'Hôpital's to ensure they are appropriate.

Addressing indeterminate forms correctly is crucial in problem-solving since it often uncovers well-defined numbers from seemingly undefined or infinite results.

Derivatives

Derivatives are a fundamental tool in calculus that describe how a function changes at a particular point. They signify the rate of change or the slope of the function at any given moment.

In the context of our problem, we encounter derivatives when applying l'Hôpital's Rule. We needed:

• The derivative of the denominator \( x \), which is straightforward as it's constant and equal to \( 1 \).
• The derivative of the numerator, found using the Fundamental Theorem of Calculus. Here, the derivative of the integral \( \int_{0}^{x} \sqrt{1+\sin t} \, dt \) is simply \( \sqrt{1+\sin x} \).

By calculating these derivatives, we transformed the original limit problem into a much simpler form that can be easily evaluated. Understanding derivatives helps significantly as they form the crux of many higher mathematics applications, especially in the study of motion, growth, and change processes.

Integrals

Integrals are the cornerstone of calculus, representing the accumulation of quantities and often used to calculate areas under curves. They come in two primary forms: indefinite, which generalize functions, and definite, which calculate actual numerical values.

In the presented exercise, the integral \( \int_{0}^{x} \sqrt{1+\sin t} \, dt \) is evaluated as \( x \) approaches zero. Here, this integral represents the total area from 0 up to \( x \), accumulated under the curve of \( \sqrt{1+\sin t} \).

• This integral involved is directly used to find its derivative through the Fundamental Theorem of Calculus for the application of l'Hôpital's Rule.
• Integrals can be complex, but they are also powerful means to work with continuous average quantities, such as areas, volumes, and other physical quantities. They help transform endless sums into manageable calculations.

Essentially, mastering integrals allows us to deeply engage with physical problems and their solutions, enabling us to predict and compute outcomes over real-world applications.