Problem 49 - Calculus

Question

Find \(\lim _{x \rightarrow 0} \frac{1}{x} \int_{0}^{x} \frac{1+t}{2+t} d t\)

Step-by-Step Solution

Answer

The limit is \( \frac{1}{2} \).

Step 1: Understand the limit problem

We are asked to find the limit \( \lim _{x \rightarrow 0} \frac{1}{x} \int_{0}^{x} \frac{1+t}{2+t} dt \). This expression involves a limit, an integral, and division by \( x \). It's important to recognize that this problem suggests using L'Hôpital's Rule, since both the numerator and the denominator approach 0 as \( x \rightarrow 0 \).

Step 2: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, the derivative of an integral \( \int_{a}^{x} f(t) dt \) is \( f(x) \). This means \( \frac{d}{dx} \left( \int_{0}^{x} \frac{1+t}{2+t} dt \right) = \frac{1+x}{2+x} \).

Step 3: Use L'Hôpital's Rule

Since both the numerator \( \int_{0}^{x} \frac{1+t}{2+t} dt \) and denominator \( x \) approach 0 as \( x \rightarrow 0 \), we can apply L'Hôpital's Rule. This allows us to take the derivative of the numerator and the derivative of the denominator separately. We already found that the derivative of the numerator is \( \frac{1+x}{2+x} \) and the derivative of the denominator is \( 1 \).

Step 4: Evaluate the Limit

Now we evaluate the limit of the resulting expression after applying L'Hôpital's Rule: \[ \lim _{x \rightarrow 0} \frac{1+x}{2+x}. \]Substituting \( x = 0 \) directly, we get\[\frac{1+0}{2+0} = \frac{1}{2}. \]

Step 5: Conclusion

Thus, the limit of the given expression as \( x \rightarrow 0 \) is \( \frac{1}{2} \). This confirms that some seemingly complex limits can be simplified using fundamental calculus techniques.

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a practical tool in calculus to tackle limits that present indeterminate forms, typically in the format \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It provides a way to find the limit of a quotient by taking the derivative of the numerator and the derivative of the denominator separately.

• Initially, check if the limit results in an indeterminate form.
• Apply the rule by differentiating both the numerator and the denominator.
• Reevaluate the limit after differentiation.

L'Hôpital's Rule is quite essential in calculus since it simplifies complex limits by transforming them into manageable expressions. In the exercise, the expression \( \frac{1}{x} \int_{0}^{x} \frac{1+t}{2+t} dt \) can be expressed in the \( \frac{0}{0} \) form as \( x \to 0 \), allowing us to apply L'Hôpital's Rule effectively. By doing so, we turn this daunting limit problem into a much more manageable task.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone concept connecting differentiation and integration. It states that if you have an integral of a function from a constant to \( x \), the derivative of that integral will yield the original function.

• The theorem bridges the concept of antiderivatives and integrals.
• It helps with computing integrals more easily through differentiation.

In the given exercise, applying this theorem simplifies the process greatly. By evaluating the integral \( \int_{0}^{x} \frac{1+t}{2+t} dt \), we find its derivative to be \( \frac{1+x}{2+x} \). This form is critical as it aids in simplifying the limit through the differentiation process and plays a pivotal role in solving the problem efficiently.

Limits

Limits are fundamental in calculus, allowing us to explore the behavior of functions as they approach specific points or infinity. A limit identifies the value that a function "approaches" as the input gets closer to a given point.

• Determine the behavior of a function near a point rather than at the point.
• Explore continuity, derivatives, and integrals.

In the problem, we examine the limit \( \lim _{x \rightarrow 0} \frac{1}{x} \int_{0}^{x} \frac{1+t}{2+t} dt \). The practice of solving this involves understanding how to simplify expressions approaching zero and modifying the function to evaluate it more clearly. Limits are crucial steps in resolving calculus problems like integrals and can sometimes lead to more elegant solutions in originally complex scenarios.

Integration

Integration is the process of finding the integral of a function, which essentially is the reverse process of differentiation. It helps in calculating areas under curves, among other applications.

• Converts the rate of change back into the original quantity.
• Two main types: definite and indefinite integrals.

In solving the exercise, integration finds the cumulative area under the function \( \frac{1+t}{2+t} \) from \( 0 \) to \( x \). Integrals like these show how a function accumulates over an interval, and in this specific problem, integrating initially sets up the application of both L'Hôpital's Rule and the Fundamental Theorem of Calculus to reduce complexity. Understanding integration is key in calculus, enabling a translation between rates of change and total quantities, critical in many real-world applications.