Problem 14 - Single Variable Calculus

Question

evaluate the limit using l'Hôpital's Rule if appropriate. $$ \lim _{u \rightarrow \pi} \frac{2 \sin ^{2} u}{1+\cos u} $$

Step-by-Step Solution

Answer

The short answer for the given limit is: \[ \lim _{u \rightarrow \pi} \frac{2 \sin ^{2} u}{1+\cos u} = 2 \]

Step 1: Identify Indeterminate Form

First, we need to verify if the given limit is in an indeterminate form. Let's check by plugging in \(u = \pi\): \[ \frac{2 \sin ^{2} (\pi)}{1+\cos (\pi)} \] Since \(\sin^2(\pi) = 0\) and \(\cos(\pi) = -1\), we get 0/0, which is an indeterminate form. As this is an indeterminate form, we can proceed with l'Hôpital's Rule.

Step 2: Applying l'Hôpital's Rule

To apply l'Hôpital's Rule, we need to differentiate the numerator and denominator with respect to \(u\) and then find the limit. Differentiating the numerator, using the chain rule and the double angle formula: \[ \frac{d}{d u} [2 \sin ^{2} u] = \frac{d}{d u} [2 \cdot \frac{1 - \cos(2u)}{2} ] = -\sin(2u) \] Differentiating the denominator: \[ \frac{d}{d u} [1+\cos u] = -\sin u \] Now, we have: \[ \lim _{u \rightarrow \pi} \frac{-\sin(2u)}{-\sin u} \] By applying l'Hôpital's Rule, we get: \[ \lim _{u \rightarrow \pi} \frac{-\sin(2u)}{-\sin u} \]

Step 3: Evaluate the Limit

Now we can directly evaluate this limit by plugging in \(u = \pi\): \[ \frac{-\sin(2(\pi))}{-\sin (\pi)} \] Since \(\sin(2\pi) = 0\) and \(\sin(\pi) = 0\), the limit is still in an indeterminate form. So, we need to apply l'Hôpital's Rule once more. Differentiating the numerator: \(\frac{d}{d u} [-\sin(2u)] = -2\cos(2u)\) Differentiating the denominator: \(\frac{d}{d u} [-\sin u] = -\cos u\) The new limit is: \[ \lim _{u \rightarrow \pi} \frac{-2\cos(2u)}{-\cos u} \] Evaluating this limit, we have: \[ \frac{-2\cos(2(\pi))}{-\cos (\pi)} = \frac{-2\cos(2\pi)}{\cos(\pi)} \] Since \(\cos(2\pi) = 1\) and \(\cos(\pi) = -1\), we get the limit as follows: \[ \frac{-2(1)}{-1} = 2 \] Therefore, the limit is: \[ \lim _{u \rightarrow \pi} \frac{2 \sin ^{2} u}{1+\cos u} = 2 \]

Key Concepts

Indeterminate Form

When we solve limits, encountering an indeterminate form can be confusing. An indeterminate form like \(\frac{0}{0}\) occurs when attempting to directly substitute a value into a function results in an undefined expression. In our problem, by evaluating the limit as \(u\) approaches \(\pi\), we obtained \(\frac{0}{0}\). This indicates that a direct evaluation isn't possible. Instead of quitting, finding this form tells us that l'Hôpital's Rule might be necessary. Recognizing an indeterminate form acts as a red flag that we need other strategies, like differentiating both the numerator and denominator, to find a solvable form.

Limit Evaluation

Evaluating limits is fundamental in calculus, helping us understand how a function behaves around a point. The goal is to find what value a function approaches as the variable gets infinitely close to a specific number. When direct evaluation gives an indeterminate form, like \(\frac{0}{0}\) in our exercise, it's a sign to use more advanced tools.

• Start by substituting the value into the function if possible.
• If an indeterminate form emerges, consider using l'Hôpital's Rule.

l'Hôpital's Rule involves finding the derivative of both the numerator and the denominator until the limit can be directly evaluated without resulting in an indeterminate form.

Differentiation

Differentiation is the process of finding the derivative, which represents the rate of change or the slope of a function. In limit evaluation via l'Hôpital's Rule, differentiation is key. For the given function \(\frac{2\sin^2 u}{1+\cos u}\), we determine the derivatives of both parts to avoid the indeterminate form directly:

• The derivative of the numerator involves using trigonometric identities and rules, such as the chain rule and double angle identities.
• The numerator simplifies to \(-\sin(2u)\), while the denominator to \(-\sin u\).

Once differentiated, a new limit needs to be evaluated. The process may require repetitive application of differentiation until solvable.

Trigonometric Functions

Trigonometric functions, like sine and cosine, are crucial in calculus for functions involving angles. In limits with trigonometric terms, understanding identities and properties aids in simplifying expressions.

• Sine and cosine functions oscillate between -1 and 1, which affects their limits.
• Our problem used identities like \(\sin^2(u) = \frac{1 - \cos(2u)}{2}\), simplifying derivatives.

When applying l'Hôpital's Rule, recognizing these identities helps compute derivatives accurately and simplifies otherwise complex expressions, making limits more approachable.