AP Calculus BC 4.7 Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms Study Notes - New Syllabus
AP Calculus BC 4.7 Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms Study Notes- New syllabus
AP Calculus BC 4.7 Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- L’Hospital’s Rule allows us to determine the limits of some indeterminate forms.
Key Concepts:
- Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms
L’Hospital’s Rule for Determining Limits of Indeterminate Forms
L’Hospital’s Rule for Determining Limits of Indeterminate Forms
L’Hospital’s Rule is a technique used in calculus to evaluate limits of indeterminate forms such as:
- \( \frac{0}{0} \)
- \( \frac{\infty}{\infty} \)
It is based on the derivatives of the numerator and denominator and can simplify difficult limits into solvable forms.
Conditions for L’Hospital’s Rule:
- The original limit must produce an indeterminate form: \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
- The functions in the numerator and denominator must be differentiable near the point of interest.
- Apply the rule: \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \quad \text{(if limit of RHS exists)} \)
Example:
Evaluate \( \displaystyle \lim_{x \to 0} \frac{\sin x}{x} \)
▶️Answer/Explanation
As \( x \to 0 \), both the numerator and denominator approach 0: \( \frac{\sin x}{x} \to \frac{0}{0} \Rightarrow \text{indeterminate form}. \)
Apply L’Hospital’s Rule:
- Differentiate numerator: \( \frac{d}{dx}(\sin x) = \cos x \)
- Differentiate denominator: \( \frac{d}{dx}(x) = 1 \)
Now take the limit: \( \lim_{x \to 0} \frac{\cos x}{1} = \cos(0) = 1 \)
Example:
Evaluate \( \displaystyle \lim_{x \to \infty} \frac{2x^2 + 5x}{3x^2 – x} \)
▶️Answer/Explanation
As \( x \to \infty \), numerator and denominator both → ∞: \( \frac{2x^2 + 5x}{3x^2 – x} \to \frac{\infty}{\infty} \Rightarrow \text{indeterminate form}. \)
Apply L’Hospital’s Rule:
- Differentiate numerator: \( \frac{d}{dx}(2x^2 + 5x) = 4x + 5 \)
- Differentiate denominator: \( \frac{d}{dx}(3x^2 – x) = 6x – 1 \)
Now evaluate: \( \lim_{x \to \infty} \frac{4x + 5}{6x – 1} = \lim_{x \to \infty} \frac{4 + \frac{5}{x}}{6 – \frac{1}{x}} = \frac{4}{6} = \frac{2}{3} \)
Other Indeterminate Forms & How to Handle Them
Not all indeterminate forms are simple quotients. Some require algebraic manipulation before L’Hospital’s Rule can be applied:
| Indeterminate Form | Conversion Method |
|---|---|
| \( 0 \cdot \infty \) | Rewrite as a quotient, e.g., \( a(x) \cdot b(x) = \frac{a(x)}{1/b(x)} \) or \( \frac{b(x)}{1/a(x)} \) |
| \( \infty – \infty \) | Combine expressions into a single fraction |
| \( 1^\infty \), \( 0^0 \), \( \infty^0 \) | Take natural logarithm, use L’Hospital’s on the exponent, then exponentiate result |
Example:
Evaluate \( \displaystyle \lim_{x \to 0^+} x \ln x \)
▶️Answer/Explanation
As \( x \to 0^+ \):
- \( x \to 0 \)
- \( \ln x \to -\infty \)
So, we have \( 0 \cdot (-\infty) \) → indeterminate form.
Rewrite as: \( x \ln x = \frac{\ln x}{1/x} \) Now we have \( \frac{-\infty}{\infty} \), still indeterminate.
Apply L’Hospital’s Rule: \( \lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} -x = 0 \)
Example:
Evaluate \( \displaystyle \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x \)
▶️Answer/Explanation
This is an indeterminate form of the type \( 1^\infty \).
Let: \( y = \left(1 + \frac{1}{x}\right)^x \Rightarrow \ln y = x \ln \left(1 + \frac{1}{x} \right) \)
Now evaluate: \( \lim_{x \to \infty} \ln y = \lim_{x \to \infty} x \ln \left(1 + \frac{1}{x} \right) \Rightarrow \text{indeterminate: } \infty \cdot 0 \)
Convert and apply L’Hospital’s Rule: \( \lim_{x \to \infty} \frac{\ln \left(1 + \frac{1}{x} \right)}{1/x} \) Differentiate numerator and denominator: \( \frac{\frac{-1}{x^2(1 + \frac{1}{x})}}{-1/x^2} = \frac{1}{1 + \frac{1}{x}} \to 1 \) So, \( \lim_{x \to \infty} \ln y = 1 \Rightarrow y = e^1 = e \)
Example:
Evaluate the following limit:
\( \displaystyle \lim_{x \to \infty} \left( \sqrt{x^2 + x} – x \right) \)
Which of the following is the correct value of the limit?
- \( \infty \)
- \( 0 \)
- \( \dfrac{1}{2} \)
- \( 1 \)
- Does not exist
▶️Answer/Explanation
This is an indeterminate form of the type \( \infty – \infty \):
\( \sqrt{x^2 + x} – x \approx \text{both terms go to } \infty \)
To resolve this, we rationalize the expression:
\( \sqrt{x^2 + x} – x = \frac{(\sqrt{x^2 + x} – x)(\sqrt{x^2 + x} + x)}{\sqrt{x^2 + x} + x} = \frac{x^2 + x – x^2}{\sqrt{x^2 + x} + x} = \frac{x}{\sqrt{x^2 + x} + x} \)
Now simplify numerator and denominator as \( x \to \infty \):
Factor out \( x \):
\( \frac{x}{\sqrt{x^2(1 + \frac{1}{x})} + x} = \frac{x}{x\sqrt{1 + \frac{1}{x}} + x} = \frac{1}{\sqrt{1 + \frac{1}{x}} + 1} \to \frac{1}{1 + 1} = \frac{1}{2} \)
Correct Answer: (C) \( \dfrac{1}{2} \)