Problem 95 - Calculus Single Variable

Question

Illustrate the assertion of l'Hôpital's Rule by plotting \(f / g\) and \(f^{\prime} / g^{\prime}\) in an interval centered at \(c\) \(f(x)=(1+\cos (x)), g(x)=(x-\pi) \sin (x), c=\pi\)

Step-by-Step Solution

Answer

Graph \(\frac{1+\cos(x)}{(x-\pi)\sin(x)}\) and \(\frac{-\sin(x)}{(x-\pi)\cos(x) + \sin(x)}\) around \(x=\pi\); both converge at \(x=\pi\), illustrating l'Hôpital's Rule.

Step 1: Understanding l'Hôpital's Rule

In problems where you encounter forms like \(\frac{0}{0}\), l'Hôpital's Rule helps by stating: \(\lim_{{x \to c}} \frac{f(x)}{g(x)} = \lim_{{x \to c}} \frac{f'(x)}{g'(x)}\), provided the right-hand limit exists. We will use this rule for \(f(x) = 1 + \cos(x)\) and \(g(x) = (x-\pi)\sin(x)\) at \(c = \pi\).

Step 2: Evaluate Original Functions at \(x = \pi\)

Calculate \(f(\pi)\) and \(g(\pi)\): \(f(\pi) = 1 + \cos(\pi) = 0\) and \(g(\pi) = (\pi - \pi)\sin(\pi) = 0\). Both are 0, indicating \(\frac{0}{0}\) form at \(x = \pi\). This is why l'Hôpital's Rule applies.

Step 3: Compute Derivatives

Find the derivatives: \(f'(x) = -\sin(x)\) and \(g'(x) = (x - \pi)\cos(x) + \sin(x)\). We differentiate \(g(x)\) using the product rule: \((u v)' = u'v + uv'\) with \(u = x-\pi\) and \(v = \sin(x)\).

Step 4: Evaluate Derivatives at \(x = \pi\)

Calculate \(f'(\pi)\) and \(g'(\pi)\): \(f'(\pi) = -\sin(\pi) = 0\) and \(g'(\pi) = 0 \cdot \cos(\pi) + 1 \cdot (-1) = -1\). The denominator does not equal zero, allowing application of l'Hôpital's Rule.

Step 5: Plot \(\frac{f(x)}{g(x)}\) and \(\frac{f'(x)}{g'(x)}\)

Graph both \(\frac{1 + \cos(x)}{(x-\pi)\sin(x)}\) and \(\frac{-\sin(x)}{(x-\pi)\cos(x) + \sin(x)}\) around \(x = \pi\). Both graphs should demonstrate the convergence around \(x = \pi\), illustrating \(\lim_{{x \to \pi}} \frac{f(x)}{g(x)} = \lim_{{x \to \pi}} \frac{f'(x)}{g'(x)}\).

Step 6: Verify the Solution

Upon plotting, verify that both \(\frac{f(x)}{g(x)}\) and \(\frac{f'(x)}{g'(x)}\) converge to the same value at \(x = \pi\). This should affirm l'Hôpital's Rule, showing the derivative ratios equate to the original limit where the function initially indicated an indeterminate form.

Key Concepts

Indeterminate Forms

When calculating limits, you may often encounter forms known as indeterminate forms. These typically arise as a result of expressions that don't give clear information about their limits, such as the form \(\frac{0}{0}\).
This means both the numerator and the denominator of the fractional expression evaluate to zero at a particular point in the domain, making it impossible to simply evaluate the limit by direct substitution.
Other indeterminate forms besides \(\frac{0}{0}\) include \(\frac{\infty}{\infty}\), \(0 \cdot \infty\), and differences like \(\infty - \infty\).

In such cases, special approaches like l'Hôpital's Rule can be applied to find the limit, assuming certain conditions are met. The application of l'Hôpital's Rule in our example allows for further evaluation of expressions at \(x = \pi\) for the functions \(f(x) = 1 + \cos(x)\) and \(g(x) = (x-\pi)\sin(x)\).
Recognizing an indeterminate form is crucial for choosing the right mathematical tool for limit evaluation.

Derivatives

Derivatives are fundamental in calculus, providing the rate at which a function changes. For a given function \(f(x)\), its derivative \(f'(x)\) tells us how fast \(f(x)\) changes with respect to changes in \(x\).
In this exercise, we need to find the derivatives of the functions \(f(x) = 1 + \cos(x)\) and \(g(x) = (x-\pi)\sin(x)\).

- The derivative of \(f(x)\) is \(f'(x) = -\sin(x)\).
- For \(g(x)\), we use the product rule since it is a product of two functions: \(u = x-\pi\) and \(v = \sin(x)\).
The product rule states that \((uv)' = u'v + uv'\), resulting in \(g'(x) = \cos(x)(x-\pi) + \sin(x)\).

Understanding how to compute these derivatives and applying rules like the product rule is essential for using techniques such as l'Hôpital’s Rule.

Limit Calculation

Calculating limits involves determining the value that a function approaches as \(x\) gets closer to a particular point. In this context, it’s about finding \(\lim_{{x \to \pi}} \frac{f(x)}{g(x)}\).
This particular limit initially undergoes a \(\frac{0}{0}\) indeterminate form, suggesting we can apply l'Hôpital's Rule to find an equivalent expression involving derivatives.

The key is first recognizing the condition for indeterminate form and then calculating the derivatives \(f'(x)\) and \(g'(x)\).
At \(x = \pi\), both \(f'(x)\) and \(g'(x)\) are calculated as \(f'\!(\pi) = 0\) and \(g'\!(\pi) = -1\).
The denominator of the derivative ratio is non-zero, ensuring that l'Hôpital's Rule applies effectively in evaluating the limit.
By determining \(\lim_{{x \to \pi}} \frac{f'(x)}{g'(x)} = \lim_{{x \to \pi}} \frac{0}{-1} = 0\), the convergence of the graphs confirms the solution.

Product Rule

The product rule is a handy tool in differentiation when dealing with the product of two functions. Calculating the derivative of a product requires considering how both functions change.
For functions \(u(x)\) and \(v(x)\), the product rule is expressed as \((u v)' = u'v + uv'\).
Applying this to \(g(x) = (x-\pi)\sin(x)\), we let \(u = x-\pi\) and \(v = \sin(x)\).
- \(u' = 1\)
- \(v' = \cos(x)\)

Thus, \((x-\pi)\)'\(\sin(x) + (x-\pi)\cos(x)\) which simplifies to \(g'(x) = \sin(x) + (x-\pi)\cos(x)\).
The product rule is crucial in evaluating complex derivatives like \(g'(x)\), ensuring an accurate computation for applying techniques like l'Hôpital's Rule.