Question

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

Short answer

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

Step-by-step solution

Understanding the Limit

The limit we need to evaluate is \( \lim _{x \rightarrow 0} \frac{1}{x} \int_{0}^{x} \frac{1+t}{2+t} \, dt \). This expression involves both a definite integral from 0 to \( x \) and a limit as \( x \) approaches 0.

Recognizing the Application of Fundamental Theorem of Calculus

Notice that the structure \( \frac{1}{x} \int_{0}^{x} f(t) \, dt \) is suggestive of the Fundamental Theorem of Calculus concerning the average value of the function \( f(t) \) over the interval \([0, x]\). This implies that as \( x \to 0 \), we are interested in \( f(0) \).

Setting Up the Expression Inside the Limit

Express the problem in terms of average value: \( \lim _{x \to 0} \frac{1}{x} \int_{0}^{x} \frac{1+t}{2+t} \, dt = \lim _{x \to 0} \left( \text{Average value of} \frac{1+t}{2+t} \text{ on } [0, x] \right) \).

Applying the L'Hôpital's Rule

To solve the given limit, notice that it is in the form of 0/0 as \( x \to 0 \). Apply L'Hôpital's rule, which gives: \( \lim _{x \to 0} \frac{G(x)}{x} = G'(0) \) if \( \lim _{x \to 0} G(x) = 0 \). Here, \( G(x) = \int_{0}^{x} \frac{1+t}{2+t} \, dt \). Therefore, we find \( G'(x) = \frac{1+x}{2+x} \).

Evaluating the Derivative at Zero

Evaluate \( G'(x) = \frac{1+x}{2+x} \) at \( x = 0 \): this is \( G'(0) = \frac{1+0}{2+0} = \frac{1}{2} \).

Calculating the Limit

Thus, we have \( \lim _{x \to 0} \frac{1}{x} \int_{0}^{x} \frac{1+t}{2+t} \, dt = G'(0) = \frac{1}{2} \).

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation with integration. It's like a bridge between finding areas under curves and determining how a function changes at any point. This theorem states that if you have a function that's continuous over some interval, then the integral of the function over that interval can be "undone" by taking the derivative of the integral.
This is particularly useful when you're dealing with limits of integrals, like in our original exercise. When you see an expression like \( \frac{1}{x} \int_{0}^{x} f(t) \, dt \), as \( x \to 0 \), the theorem helps you convert the integral problem into a simpler differentiation problem.

• If you have a continuous function \( f(t) \), the integral from 0 to \( x \) can be represented as \( G(x) = \int_{0}^{x} f(t) \, dt \).
• By the theorem, the derivative \( G'(x) \) can be found by simply evaluating \( f(x) \).

By applying this theorem, complex problems related to limits involving integrals become manageable, as derivatives are often easier to evaluate.

average value of a function

The average value of a function over an interval helps to characterize the function's behavior, providing a single value that summarizes its overall performance across the interval. It's like finding the mean height if you measured various points on a bumpy road. The formula to determine the average value of a continuous function \( f(x) \) over \([a, b]\) is:
\[\frac{1}{b-a} \int_a^b f(x) \, dx\]
This formula calculates the total area under \( f(x) \) over \([a, b]\), and then it divides by the length of that interval.
In the exercise, we deal with the interval \([0, x]\), meaning the average value of the function \( \frac{1+t}{2+t} \) approaches the value of the function at 0, as \( x \to 0 \).

• Think of \( x \to 0 \) as the interval shrinking to a single point, focusing on the behavior of the function precisely at \( t = 0 \).
• This illustrates the value of the function's average on infinitely small intervals, linking back to the Fundamental Theorem's concept of uncovering values via limits and derivatives.

This approach simplifies understanding and solving complex limit problems involving integrals.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus used to solve limits that result in indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). To utilize L'Hôpital's Rule, both the numerator and the denominator must be differentiable near the point of interest, and you should verify that the limit exists after applying the rule.
In simple terms, if you find such a limit:\[\lim_{x \to c} \frac{f(x)}{g(x)}\]
Results in an indeterminate form, then you can find this limit by calculating:\[\lim_{x \to c} \frac{f'(x)}{g'(x)}\]
provided this new form is no longer indeterminate.

• In our problem, we had \( \frac{1}{x} \int_{0}^{x} \frac{1+t}{2+t} \, dt \), leading to the form \( \frac{0}{0} \) as \( x \to 0 \).
• L'Hôpital's Rule allowed us to transform this into a simple derivative evaluation at 0, revealing the limit as \( \frac{1}{2} \).

L'Hôpital's Rule simplifies finding tricky limits by converting them into mere differentiation tasks, making it a vital strategy in calculus for tackling complex limit calculations.