#AP Calculus AB/BC: L'Hôpital's Rule - Your Ultimate Guide 🚀

Hey there, future AP Calculus master! Let's dive into L'Hôpital's Rule, a super handy tool for tackling those tricky indeterminate limits. This guide is designed to be your go-to resource, especially the night before the exam. Let's make sure you're feeling confident and ready! 💪

#4.7: L'Hôpital's Rule for Indeterminate Forms

Remember those limits that gave us $\frac{0}{0}$ or $\pm\frac{\infty}{\infty}$? Those are called indeterminate forms, and they're where L'Hôpital's Rule shines. Instead of algebraic manipulations, we get to use derivatives! 🥳

#📏 What is L'Hôpital's Rule?

#✏️ L'Hôpital's Rule: Step-by-Step

Let's walk through an example together. Evaluate:

$$\lim_{x \to \frac{\pi}{2}}\frac{\cos(x)}{x-\frac{\pi}{2}}$$

1. Check for Indeterminate Form: Plugging in $x = \frac{\pi}{2}$ gives us $\frac{0}{0}$. ✅

2. Verify Conditions (FRQ Must-Do):

   $$\lim_{x \to \frac{\pi}{2}}\cos(x) = 0$$

   $$\lim_{x \to \frac{\pi}{2}}x - \frac{\pi}{2} = 0$$

   Since both limits are 0, we can apply L'Hôpital's Rule. (Make sure to state this in the exam!)

3. Apply L'Hôpital's Rule:

   $$\lim_{x \to \frac{\pi}{2}}\frac{\cos(x)}{x-\frac{\pi}{2}} = \lim_{x\to \frac{\pi}{2}}\frac{\frac{d}{dx}[\cos(x)]}{\frac{d}{dx}[x-\frac{\pi}{2}]}$$

4. Take Derivatives:

   $$\lim_{x\to \frac{\pi}{2}}\frac{-\sin(x)}{1}$$

5. Evaluate the Limit:

   $$= \frac{-\sin(\frac{\pi}{2})}{1} = -1$$

So, $\lim_{x \to \frac{\pi}{2}}\frac{\cos(x)}{x-\frac{\pi}{2}} = -1$. 🎉

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#📝 Practice Time!

Ready to try it yourself? Here are a couple of practice problems. Treat them like FRQs and show all your steps!

#❓ L'Hôpital's Rule: Practice Problems

#Question 1:

$$\lim_{x\to 0}\frac{\tan(x)}{7x+\tan(x)}$$

#Question 2:

$$\lim_{x\to \infty}\frac{3x^2-8}{7x^2+21}$$

#✅ L'Hôpital's Rule: Solutions

#Question 1:

1. Indeterminate Form: Plugging in $x=0$ gives $\frac{0}{0}$.

2. Verify Conditions:

   $$\lim_{x\to 0}\tan(x) = 0$$

   $$\lim_{x\to 0}7x + \tan(x) = 0$$

3. Apply L'Hôpital's Rule:

   $$\lim_{x\to 0}\frac{\tan(x)}{7x+\tan(x)} = \lim_{x\to 0}\frac{\frac{d}{dx}[\tan(x)]}{\frac{d}{dx}[7x+\tan(x)]}$$

4. Take Derivatives:

   $$\lim_{x\to 0}\frac{\sec^2(x)}{7+\sec^2(x)}$$

5. Evaluate the Limit:

   $$= \frac{\sec^2(0)}{7+\sec^2(0)} = \frac{1}{7+1} = \frac{1}{8}$$

   Thus, $\lim_{x\to 0}\frac{\tan(x)}{7x+\tan(x)} = \frac{1}{8}$.

#Question 2:

1. Indeterminate Form: Plugging in $x = \infty$ gives $\frac{\infty}{\infty}$.

2. Verify Conditions:

   $$\lim_{x\to \infty}3x^2 - 8 = \infty$$

   $$\lim_{x\to \infty}7x^2 + 21 = \infty$$

3. Apply L'Hôpital's Rule:

   $$\lim_{x\to \infty}\frac{3x^2-8}{7x^2+21} = \lim_{x\to \infty}\frac{\frac{d}{dx}[3x^2-8]}{\frac{d}{dx}[7x^2+21]}$$

4. Take Derivatives:

   $$\lim_{x\to \infty}\frac{6x}{14x}$$

5. Simplify and Evaluate:

   $$= \frac{6}{14} = \frac{3}{7}$$

   Therefore, $\lim_{x\to \infty}\frac{3x^2-8}{7x^2+21} = \frac{3}{7}$.

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#💫 Final Exam Focus

Key Takeaways for Exam Day:

• Indeterminate Forms: Always check for $\frac{0}{0}$ or $\pm\frac{\infty}{\infty}$ before using L'Hôpital's Rule.
• FRQ Steps: Show the limits of the numerator and denominator separately. State that you're applying L'Hôpital's Rule.
• Derivatives: Make sure you know your derivative rules inside and out! (Trig, polynomials, exponentials, etc.)
• Simplification: Simplify after taking derivatives. Sometimes you might need to apply L'Hôpital's Rule multiple times.
• Connections: L'Hôpital's Rule often appears in combination with other concepts, so be ready to use it in different contexts.

#

Practice Question

Practice Questions

#Multiple Choice Questions

1. Evaluate $\lim_{x \to 0} \frac{\sin(3x)}{x}$: (A) 0 (B) 1 (C) 3 (D) Does not exist

2. Evaluate $\lim_{x \to \infty} \frac{2x^2 + 5x}{3x^2 - 1}$: (A) 0 (B) 2/3 (C) 5/1 (D) Does not exist

3. Evaluate $\lim_{x \to 1} \frac{\ln(x)}{x-1}$: (A) 0 (B) 1 (C) -1 (D) Does not exist

#Free Response Question

Consider the function $f(x) = \frac{e^x - 1 - x}{x^2}$.

(a) Show that $\lim_{x \to 0} f(x)$ is an indeterminate form. (2 points)

(b) Evaluate $\lim_{x \to 0} f(x)$. (4 points)

(c) Let $g(x) = \frac{e^x - 1 - x}{x}$. Evaluate $\lim_{x \to 0} g(x)$. (3 points)

#FRQ Scoring Rubric

(a)

• 1 point: Shows $\lim_{x \to 0} (e^x - 1 - x) = 0$
• 1 point: Shows $\lim_{x \to 0} x^2 = 0$

(b)

• 1 point: Applies L'Hôpital's Rule once correctly
• 1 point: Takes the derivatives correctly
• 1 point: Applies L'Hôpital's Rule again correctly
• 1 point: Evaluates the limit correctly

(c)

• 1 point: Shows $\lim_{x \to 0} (e^x - 1 - x) = 0$ and $\lim_{x \to 0} x = 0$
• 1 point: Applies L'Hôpital's Rule correctly
• 1 point: Evaluates the limit correctly

#Solutions to Practice Questions

#Multiple Choice

1. (C) 3 (Apply L'Hôpital's Rule)
2. (B) 2/3 (Apply L'Hôpital's Rule or compare leading coefficients)
3. (B) 1 (Apply L'Hôpital's Rule)

#Free Response Question

(a)

$$\lim_{x \to 0} (e^x - 1 - x) = e^0 - 1 - 0 = 1 - 1 = 0$$

$$\lim_{x \to 0} x^2 = 0^2 = 0$$

Since both limits are 0, the limit is of the form $\frac{0}{0}$, which is an indeterminate form.

(b)

Applying L'Hôpital's Rule once:

$$\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \lim_{x \to 0} \frac{e^x - 1}{2x}$$

This is still an indeterminate form $\frac{0}{0}$, so apply L'Hôpital's Rule again:

$$\lim_{x \to 0} \frac{e^x - 1}{2x} = \lim_{x \to 0} \frac{e^x}{2} = \frac{e^0}{2} = \frac{1}{2}$$

Thus, $\lim_{x \to 0} f(x) = \frac{1}{2}$.

(c)

$$\lim_{x \to 0} (e^x - 1 - x) = 0$$

$$\lim_{x \to 0} x = 0$$

Applying L'Hôpital's Rule:

$$\lim_{x \to 0} \frac{e^x - 1 - x}{x} = \lim_{x \to 0} \frac{e^x - 1}{1} = e^0 - 1 = 1 - 1 = 0$$

Thus, $\lim_{x \to 0} g(x) = 0$.

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You've got this! Keep practicing, stay calm, and remember all the cool tricks you've learned. You're going to rock this exam! 🌟