Welcome to Section 4.4: Indeterminate Forms and L'Hôpital's Rule!
In this section, we're going to explore a powerful technique for evaluating limits that initially appear undefined. These limits involve what we call indeterminate forms, and we'll use L'Hôpital's Rule to conquer them. Don't worry, it's not as intimidating as it sounds! Let's dive in!
What are Indeterminate Forms?
Sometimes, when we try to evaluate a limit by direct substitution, we encounter expressions like $\frac{0}{0}$ or $\frac{\infty}{\infty}$. These are called indeterminate forms because the value of the limit is not immediately clear. Other indeterminate forms include $0 \cdot \infty$, $\infty - \infty$, $0^0$, $\infty^0$, and $1^\infty$. They require further analysis to determine the actual limit, if it exists.
For example, consider the function:
$$F(x) = \frac{\ln x}{x - 1}$$As $x$ approaches 1, both the numerator and denominator approach 0, resulting in the indeterminate form $\frac{0}{0}$. This is where L'Hôpital's Rule comes to the rescue!
L'Hôpital's Rule: Your Limit-Solving Superhero
L'Hôpital's Rule states that if we have a limit of the form $\lim_{x \to a} \frac{f(x)}{g(x)}$ that results in an indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, and if $f$ and $g$ are differentiable with $g'(x) \neq 0$ near $a$, then:
$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$In simpler terms, we can take the derivative of the numerator and the derivative of the denominator separately and then try to evaluate the limit again. Let's look at some examples from the class notes:
Examples
- Example 1: Find $\lim_{x \to 1} \frac{\ln x}{x - 1}$.
Applying L'Hôpital's Rule:
$$\lim_{x \to 1} \frac{\ln x}{x - 1} = \lim_{x \to 1} \frac{\frac{1}{x}}{1} = \frac{\frac{1}{1}}{1} = 1$$ - Example 2: Calculate $\lim_{x \to \infty} \frac{e^x}{x^2}$.
Applying L'Hôpital's Rule twice:
$$\lim_{x \to \infty} \frac{e^x}{x^2} = \lim_{x \to \infty} \frac{e^x}{2x} = \lim_{x \to \infty} \frac{e^x}{2} = \infty$$ - Example 4: Find $\lim_{x \to 0} \frac{\tan x - x}{x^3}$.
Applying L'Hôpital's Rule multiple times:
$$\lim_{x \to 0} \frac{\tan x - x}{x^3} = \lim_{x \to 0} \frac{\sec^2 x - 1}{3x^2} = \lim_{x \to 0} \frac{2 \sec^2 x \tan x}{6x} = \lim_{x \to 0} \frac{\sec^2 x}{3} = \frac{1}{3}$$
Indeterminate Products and Differences
What about indeterminate forms like $0 \cdot \infty$ or $\infty - \infty$? The key is to rewrite the expression as a quotient so you can apply L'Hôpital's Rule. For example, if you have a limit of the form $\lim_{x \to a} f(x)g(x)$ where $f(x) \to 0$ and $g(x) \to \infty$, you can rewrite it as $\lim_{x \to a} \frac{f(x)}{\frac{1}{g(x)}}$ or $\lim_{x \to a} \frac{g(x)}{\frac{1}{f(x)}}$, which will give you either $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
Indeterminate Powers
Indeterminate powers such as $0^0$, $\infty^0$, and $1^\infty$ can be handled using logarithms. For example, to find $\lim_{x \to a} [f(x)]^{g(x)}$, let $y = [f(x)]^{g(x)}$. Then $\ln y = g(x) \ln f(x)$. Find the limit of $\ln y$ as $x \to a$, and then exponentiate the result to find the original limit.
Key Takeaways
- Identify indeterminate forms before applying any limit techniques.
- L'Hôpital's Rule is applicable for $\frac{0}{0}$ and $\frac{\infty}{\infty}$ forms.
- Rewrite expressions to fit the required form for L'Hôpital's Rule when dealing with other indeterminate forms.
- Remember to take derivatives correctly!
Keep practicing, and you'll become a master of L'Hôpital's Rule! Good luck, and see you in the next section!