L'Hôpital's Rule: Limit (7-7 Sin Θ)/(5+5 Cos 2θ)

Hey there, math enthusiasts! Ever stumbled upon a limit that just seems impossible to crack? You know, those pesky indeterminate forms like 0/0 or ∞/∞ that make you want to pull your hair out? Well, fret no more! Today, we're going to dissect a fascinating problem using one of the most powerful tools in calculus: L'Hôpital's Rule. We'll be tackling the limit ${\lim _{\theta \rightarrow \frac{\pi}{2}} \frac{7-7 \sin \theta}{5+5 \cos 2 \theta}}$, breaking it down step-by-step so you can conquer similar challenges with confidence. So, grab your calculators, sharpen your pencils, and let's dive into the world of L'Hôpital's Rule!

Understanding the Indeterminate Form

Before we jump into applying L'Hôpital's Rule, it's crucial to understand why we need it in the first place. Our initial expression is ${\lim _{\theta \rightarrow \frac{\pi}{2}} \frac{7-7 \sin \theta}{5+5 \cos 2 \theta}}$. The first thing we always do when faced with limits, guys, is to directly substitute the value that ${\theta}$ is approaching. Let's try plugging in ${\frac{\pi}{2}}$ into our expression:

Numerator: ${7 - 7 \sin(\frac{\pi}{2}) = 7 - 7(1) = 0}$

Denominator: ${5 + 5 \cos(2 \cdot \frac{\pi}{2}) = 5 + 5 \cos(\pi) = 5 + 5(-1) = 0}$

Uh oh! We've landed in the dreaded 0/0 indeterminate form. This doesn't mean the limit doesn't exist; it simply means we can't determine the limit by direct substitution alone. This is where L'Hôpital's Rule comes to our rescue! These indeterminate forms basically tell us that there's more to the limit than meets the eye initially, and we need to dig deeper to uncover the true behavior of the function as ${\theta}$ approaches ${\frac{\pi}{2}}$. Recognizing this indeterminate form is the first key step in knowing when and how to apply L'Hôpital's Rule effectively. Without this crucial observation, we might wander down incorrect paths and never find the true limit. So, always remember to check for these forms before attempting other limit-solving techniques.

Introducing L'Hôpital's Rule: Our Super Tool

Okay, so what exactly is this magical L'Hôpital's Rule everyone keeps talking about? In simple terms, it's a technique that allows us to evaluate limits of indeterminate forms by taking the derivatives of the numerator and the denominator separately. The rule states that if we have a limit of the form ${\lim_{x \rightarrow c} \frac{f(x)}{g(x)}}$, where both f(x) and g(x) approach 0 or both approach ±∞ as x approaches c, and if the limit ${\lim_{x \rightarrow c} \frac{f'(x)}{g'(x)}}$ exists, then:

${\lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \lim_{x \rightarrow c} \frac{f'(x)}{g'(x)}}$

In plain English, this means that if we have an indeterminate form, we can take the derivative of the top and the derivative of the bottom, and then try evaluating the limit again. And guess what? Sometimes, this simple trick transforms a seemingly impossible problem into a piece of cake! However, there are critical conditions that must be met before we can confidently wield this powerful rule. First, as we've already discussed, the limit must result in an indeterminate form of either 0/0 or ±∞/±∞. Attempting to apply L'Hôpital's Rule to limits that don't exhibit these forms will lead to incorrect answers. Second, both the functions f(x) and g(x) must be differentiable in an open interval containing c (except possibly at c itself). This ensures that we can actually compute the derivatives needed for the rule. Third, the limit of the derivatives, ${\lim_{x \rightarrow c} \frac{f'(x)}{g'(x)}}$, must exist. If this limit doesn't exist, L'Hôpital's Rule cannot provide a conclusive answer, and other methods may be required to evaluate the original limit. Understanding these conditions is just as important as knowing the rule itself, preventing us from misapplying it and arriving at wrong conclusions.

Applying L'Hôpital's Rule to Our Problem

Now, let's put L'Hôpital's Rule to work on our specific problem: ${\lim _{\theta \rightarrow \frac{\pi}{2}} \frac{7-7 \sin \theta}{5+5 \cos 2 \theta}}$. We've already established that direct substitution gives us the 0/0 indeterminate form, so we're good to go! The first step is to identify our f(θ) and g(θ):

f(θ) = 7 - 7sin(θ)

g(θ) = 5 + 5cos(2θ)

Next, we need to find the derivatives of both f(θ) and g(θ) with respect to θ. Remember your trig derivatives, folks! This is where your calculus toolbox comes in handy. Applying the derivative rules, we get:

f'(θ) = -7cos(θ)

g'(θ) = -10sin(2θ) (Remember the chain rule! The derivative of cos(2θ) is -2sin(2θ), and we multiply that by the 5 outside.)

Now, we apply L'Hôpital's Rule by taking the limit of the ratio of these derivatives:

${\lim _{\theta \rightarrow \frac{\pi}{2}} \frac{f'(\theta)}{g'(\theta)} = \lim _{\theta \rightarrow \frac{\pi}{2}} \frac{-7 \cos \theta}{-10 \sin 2 \theta}}$

Great! We've applied the rule once, but let's see if we can evaluate this new limit directly. Substituting θ = ${\frac{\pi}{2}}$ gives us:

${\frac{-7 \cos(\frac{\pi}{2})}{-10 \sin(2 \cdot \frac{\pi}{2})} = \frac{-7(0)}{-10 \sin(\pi)} = \frac{0}{0}}$

Blast! We're still stuck with the 0/0 indeterminate form. This isn't a setback; it's simply a sign that we might need to apply L'Hôpital's Rule again! This iterative nature of L'Hôpital's Rule is one of its key characteristics, allowing us to repeatedly differentiate until we reach a limit that can be evaluated directly. Recognizing when to reapply the rule is crucial for solving more complex problems.

Applying L'Hôpital's Rule Again (and Again!) and the Final Solution

Since we're still in the 0/0 indeterminate form, we're going to roll up our sleeves and apply L'Hôpital's Rule a second time. This means we need to find the derivatives of f'(θ) and g'(θ):

f''(θ) = 7sin(θ) (The derivative of -7cos(θ) is 7sin(θ))

g''(θ) = -20cos(2θ) (The derivative of -10sin(2θ) is -20cos(2θ))

Now, let's take the limit of the ratio of these second derivatives:

${\lim _{\theta \rightarrow \frac{\pi}{2}} \frac{f''(\theta)}{g''(\theta)} = \lim _{\theta \rightarrow \frac{\pi}{2}} \frac{7 \sin \theta}{-20 \cos 2 \theta}}$

Let's try direct substitution again. Plugging in θ = ${\frac{\pi}{2}}$, we get:

${\frac{7 \sin(\frac{\pi}{2})}{-20 \cos(2 \cdot \frac{\pi}{2})} = \frac{7(1)}{-20 \cos(\pi)} = \frac{7}{-20(-1)} = \frac{7}{20}}$

Hallelujah! We've finally arrived at a determinate form. The limit of the ratio of the second derivatives is ${\frac{7}{20}}$. According to L'Hôpital's Rule, this is also the limit of our original expression:

${\lim _{\theta \rightarrow \frac{\pi}{2}} \frac{7-7 \sin \theta}{5+5 \cos 2 \theta} = \frac{7}{20}}$

So, there you have it, guys! We successfully navigated a potentially tricky limit using the power of L'Hôpital's Rule. But our journey doesn't end here! It's essential to remember that math, like any skill, requires consistent practice. To truly master L'Hôpital's Rule, you need to tackle a variety of problems, each with its unique challenges and nuances. This hands-on experience will not only solidify your understanding of the rule itself but also sharpen your overall problem-solving abilities in calculus. Think of each problem as a puzzle, and L'Hôpital's Rule as one of the key pieces in your puzzle-solving toolkit.

Key Takeaways and Practice Makes Perfect

Let's recap the key things we've learned today:

1. Identify Indeterminate Forms: Always check for 0/0 or ∞/∞ before applying L'Hôpital's Rule.
2. Differentiate Carefully: Make sure you correctly find the derivatives of both the numerator and the denominator.
3. Reapply if Needed: Don't be afraid to apply L'Hôpital's Rule multiple times if you still encounter an indeterminate form.
4. Know the Conditions: Remember that L'Hôpital's Rule can only be used if the limit results in an indeterminate form and the derivatives exist.

To truly master L'Hôpital's Rule, practice is paramount. Seek out problems that involve trigonometric functions, exponential functions, and logarithmic functions. Experiment with limits that require multiple applications of the rule. Challenge yourself to identify when L'Hôpital's Rule is the most efficient approach and when alternative methods might be more suitable. The more you practice, the more comfortable and confident you'll become in wielding this powerful tool. And remember, guys, the journey of learning calculus is a marathon, not a sprint. Embrace the challenges, celebrate your successes, and never stop exploring the fascinating world of mathematics!