Problem 123 - Calculus for Biology and Medicine

Question

Use l'Hospital's rule to compute $$ \lim _{x \rightarrow 0} \frac{1}{x^{2}} \int_{0}^{x} \sin t d t $$

Step-by-Step Solution

Answer

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

Step 1: Understanding the Problem

We need to find the limit \[ \lim _{x \rightarrow 0} \frac{1}{x^{2}} \int_{0}^{x} \sin t \, dt \] As \( x \rightarrow 0 \), both the numerator and denominator approach 0, creating the \( \frac{0}{0} \) form, making it suitable for l'Hospital's Rule.

Step 2: Applying l'Hospital's Rule

According to l'Hospital's Rule, for \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) forms, \\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}, \] \provided the limit on the right exists. Here, \( f(x) = \int_{0}^{x} \sin t \, dt \) and \( g(x) = x^2 \).

Step 3: Differentiating the Numerator and Denominator

The derivative of the denominator \( g(x) = x^2 \) is \( g'(x) = 2x \). For the numerator, use the Fundamental Theorem of Calculus: \[ \frac{d}{dx} \left( \int_{0}^{x} \sin t \, dt \right) = \sin x. \]

Step 4: Final Limit Calculation

Substitute the derivatives back into the limit: \[ \lim _{x \rightarrow 0} \frac{\sin x}{2x}. \] This again forms \( \frac{0}{0} \), so apply l'Hospital's Rule once more: The derivative of \( \sin x \) is \( \cos x \), and for \( 2x \) it is \( 2 \). \Thus: \[ \lim _{x \rightarrow 0} \frac{\cos x}{2} = \frac{\cos(0)}{2} = \frac{1}{2}. \]

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration. It is a powerful tool because it allows us to evaluate definite integrals with ease. The theorem can be broken down into two main parts. The first part states that if you have a continuous function, say \( f(t) \), over an interval \([a, b]\), then the function defined by the integral of \( f(t) \) from \( a \) to \( x \), where \( x \) is within the interval \([a, b]\), is differentiable.
This means, if you differentiate the function \( F(x) = \int_{a}^{x} f(t) \, dt \), it simplifies beautifully to \( f(x) \). In our exercise example, the function \( f(t) = \sin t \), making the derivative \( \frac{d}{dx} \left( \int_{0}^{x} \sin t \, dt \right) \) equal to \( \sin x \).

• This simplification is the crux of applying the Fundamental Theorem.
• The theorem helps us move smoothly from the integration of a function back to a function in its derivative form.
• This step is essential when you are analyzing the behavior of integral-based expressions.

indeterminate forms

When evaluating limits, encountering indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) is quite common. These forms are termed 'indeterminate' because they do not directly lead to a specific answer, hence require further analysis. Our initial step with the limit \( \lim _{x \rightarrow 0} \frac{1}{x^{2}} \int_{0}^{x} \sin t \, dt \) leads us right into a \( \frac{0}{0} \) indeterminate form.
What does one do in such situations? Here, l'Hospital's Rule comes to the rescue. This rule provides a method to resolve these tricky indeterminate forms by focusing on derivatives.
Understanding how to transform an indeterminate form using derivatives enables us to reach a definite solution:

• Recognize the indeterminate form.
• Decide on applying l'Hospital's Rule, which involves taking the derivatives of both the numerator and denominator.
• Perform a careful analysis of derivatives during substitution to simplify and resolve the limit.

limit evaluation

Evaluating limits, especially when they result in indeterminate forms, is a fundamental skill in calculus. The limit in question, \( \lim _{x \rightarrow 0} \frac{1}{x^{2}} \int_{0}^{x} \sin t \, dt \), initially forms \( \frac{0}{0} \), making it a prime candidate for l'Hospital's Rule.
To evaluate this limit:

• Apply l'Hospital’s Rule, knowing that it’s a systematic approach for such forms.
• Differentiating the expressions, we find that \( \frac{d}{dx} \left( \int_{0}^{x} \sin t \, dt \right) = \sin x \) (from the Fundamental Theorem) and \( \frac{d}{dx}(x^2) = 2x \).
• Substitute these derivatives back to find \( \lim _{x \rightarrow 0} \frac{\sin x}{2x} \).

Since this again results in \( \frac{0}{0} \), apply l'Hospital's Rule once more by differentiating \( \sin x \) to get \( \cos x \) and \( 2x \) to get \( 2 \). Thus, the limit simplifies to \( \frac{\cos(0)}{2} = \frac{1}{2} \). This systematic approach of limit evaluation ensures clarity and correctness in solving complex calculus problems.