Problem 32 - Calculus

Question

If \(f(2)=g(2)=h(2)=0,\) and \(f^{\prime}(2)=h^{\prime}(2)=0\) \(g^{\prime}(2)=22,\) and \(f^{\prime \prime}(2)=3, g^{\prime \prime}(2)=5, h^{\prime \prime}(2)=7\) calculate the following limits. Explain your reasoning. (a) \(\lim _{x \rightarrow 2} \frac{f(x)}{h(x)}\) (b) \(\lim _{x \rightarrow 2} \frac{f(x)}{g(x)}\)

Step-by-Step Solution

Answer

(a) \(\frac{3}{7}\), (b) 0.

Step 1: Understanding the Problem Condition

Given are the functions \( f(x), g(x), h(x) \) and their derivatives around \( x = 2 \). Each function \( f, g, \) and \( h \) is zero at \( x = 2 \) \( (f(2) = g(2) = h(2) = 0)\). Additionally, \( f'(2) = h'(2) = 0 \), \( g'(2) = 22 \), \( f''(2) = 3 \), \( g''(2) = 5 \), and \( h''(2) = 7 \). These given conditions help determine the behavior of function limits at 2.

Step 2: Apply L'Hôpital's Rule to First Limit

For \( \lim_{x \to 2} \frac{f(x)}{h(x)} \), both the numerator and denominator are zero at \( x=2 \), which is an indeterminate form of type 0/0. According to L'Hôpital's Rule, we can take the derivative of the numerator and the denominator: \( \lim_{x \to 2} \frac{f(x)}{h(x)} = \lim_{x \to 2} \frac{f'(x)}{h'(x)} \). Given that \( f'(2) = h'(2) = 0 \), we encounter another 0/0. We apply L'Hôpital's Rule again: \( \lim_{x \to 2} \frac{f''(x)}{h''(x)} = \frac{3}{7} \).

Step 3: Apply L'Hôpital's Rule to Second Limit

For \( \lim_{x \to 2} \frac{f(x)}{g(x)} \), similarly, it's indeterminate form 0/0. By L'Hôpital's Rule, \( \lim_{x \to 2} \frac{f(x)}{g(x)} = \lim_{x \to 2} \frac{f'(x)}{g'(x)} \). Here, \( f'(2) = 0 \) and \( g'(2) = 22 \), leading to \( \lim_{x \to 2} \frac{0}{22} = 0 \).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a fundamental technique in calculus used to evaluate limits of indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). When the direct substitution into a limit yields these forms, the rule allows you to instead take the derivatives of the numerator and the denominator.
In our exercise, both limits involved indeterminate forms, allowing the application of this rule.

• For \( \lim_{x \to 2} \frac{f(x)}{h(x)} \), initially both the function values at \(x=2\) are 0, resulting in a \(\frac{0}{0}\) form. L'Hôpital's Rule was applied by differentiating the numerator and denominator, repeating the process until the form was no longer \(\frac{0}{0}\).
• This resulted in evaluating the second derivatives: \(\lim_{x \to 2} \frac{f''(x)}{h''(x)} \), yielding the solution \(\frac{3}{7}\).

L'Hôpital’s Rule simplifies these complex limits, making it a powerful tool when facing indeterminate forms.

Indeterminate Forms

Indeterminate forms present challenges when calculating limits. Common forms include \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\), but others like \(0 \cdot \infty\) and \(\infty - \infty\) exist too.

• Such forms imply that a straightforward evaluation at a particular point doesn’t give a clear limit value.
• For instance, \(\lim_{x \to 2} \frac{f(x)}{h(x)}\) and \(\lim_{x \to 2} \frac{f(x)}{g(x)}\) showed an initial \(\frac{0}{0}\) scenario, necessitating further steps.

Addressing these forms often involves recalculating the expressions around the point of indeterminacy, typically through differentiation, as shown in the exercise with L'Hôpital’s Rule.
Recognizing when an indeterminate form occurs is the first step in deciding which advanced calculus technique to apply, ensuring accurate limit solutions.

Derivative Applications

Derivatives enlighten us about the rate of change of functions and have vast applications, from evaluating limits to solving optimization problems. In the context of limits, derivatives help resolve indeterminate forms.

• Differentiation gives us a way to reanalyze functions near points where direct evaluation fails, like \(x=2\) in our example.
• In both limits of the exercise, derivatives of given functions were crucial: finding \(f'(x), h'(x)\), and further calculating \(f''(x), h''(x)\) for deeper insights.

Without these derivative applications, addressing limits of the given indeterminate forms would remain complex.
Thus, understanding derivatives not only deepens calculus knowledge but also equips learners to tackle a broader range of mathematical problems efficiently.