Calculus: Advanced Indeterminate Forms and L'Hôpital's Rule

Indeterminate Products (Type 2)

• Definition: An indeterminate product occurs when taking a limit of the form $\lim_{x \to a} [f(x) \times g(x)]$ where one function approaches $0$ and the other approaches $\infty$ (i.e., $0 \times \infty$).

• Contradiction of Intuition: While it might seem that multiplying anything by zero results in zero, or anything by infinity results in infinity, this is not a rule for limits. The result can be zero, infinity, or any other number.

• Conversion Strategy: To resolve this form, the product must be transformed into a quotient to apply L'Hôpital's Rule ($\frac{0}{0}$ or $\frac{\infty}{\infty}$).     * Option A: $f \times g = \frac{f}{1/g}$     * Option B: $g \times f = \frac{g}{1/f}$

• Choosing which function to "flip": This choice is not arbitrary and requires strategic thinking to avoid making the derivative more difficult.     * Hierarchical Preference for Derivatives:         * Exponentials are the easiest to deal with and should generally be moved to the denominator (by changing the sign of the exponent).         * Polynomials are easier to handle than logarithms but harder than exponentials.         * Logarithms (ln) should generally stay in the numerator because their derivatives ($\frac{1}{x}$) are easier to handle than the reciprocal of a log function.

• Example 1: Polynomial times Exponential     * Problem: $\lim_{x \to \infty} (x^3 \times e^{-x^2})$     * Initial Analysis: As $x \to \infty$, $x^3 \to \infty$ and $e^{-x^2} \to 0$. This is the indeterminate form $\infty \times 0$.     * Transformation: Use the negative exponent to bring the exponential to the denominator: $\lim_{x \to \infty} \frac{x^3}{e^{x^2}}$.     * Method: This is now $\frac{\infty}{\infty}$ form. We apply L'Hôpital's Rule.     * First Derivative: $\lim_{x \to \infty} \frac{3x^2}{2x e^{x^2}}$. Simplifying the $x$ terms results in $\lim_{x \to \infty} \frac{3x}{2e^{x^2}}$.     * Second Derivative: Apply L'Hôpital again to get $\lim_{x \to \infty} \frac{3}{4x e^{x^2}}$.     * Final Result: As the denominator goes to infinity and the numerator is a constant ($3$), the limit is $0$.

Indeterminate Powers (Type 3)

• Definition: These occur when a limit involves a function in the base and a function in the exponent, resulting in the forms:     * $0^0$     * $\infty^0$     * $1^\infty$

• Conceptual Note: $1^\infty$ is not necessarily $1$, and $0^0$ is not necessarily $0$ or $1$. These are all indeterminate.

• Method of Resolution (Logarithmic Identity): To resolve indeterminate powers, we use the natural logarithm ($\ln$) to "bring down" the exponent.     * Identity: $A^B = e^{\ln(A^B)} = e^{B \ln A}$.     * Application to limits: $\lim_{x \to a} [f(x)^{g(x)}] = e^{\lim_{x \to a} [g(x) \ln f(x)]}$.

• Two-Step Process:     1. Evaluate the limit of the exponent portion: $L = \lim_{x \to a} [g(x) \ln f(x)]$. This usually converts the power problem into a Type 2 product ($0 \times \infty$).     2. The final answer is $e^L$. It is a common mistake for students to forget this final step and provide $L$ as the answer.

• Example 1: Zero to the Power Zero     * Problem: $\lim_{x \to 0^+} x^x$.     * Form: $0^0$.     * Setup: Rewrite as $e^{\lim_{x \to 0^+} (x \ln x)}$.     * Exponent Limit: Evaluate $\lim_{x \to 0^+} (x \ln x)$. This is a $0 \times \infty$ product.         * Transform to quotient: $\lim_{x \to 0^+} \frac{\ln x}{1/x}$.         * Apply L'Hôpital: $\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0$.     * Final Result: $e^0 = 1$.

• Example 2: One to the Power Infinity     * Problem: $\lim_{x \to 0} (1 - 2x)^{1/x}$.     * Form: $1^\infty$.     * Setup: Rewrite as $e^{\lim_{x \to 0} [\frac{1}{x} \ln(1 - 2x)]}$.     * Exponent Limit: As $x \to 0$, this is an $\infty \times 0$ form that simplifies to a $\frac{0}{0}$ form: $\lim_{x \to 0} \frac{\ln(1-2x)}{x}$.

• Example 3: Infinity to the Power Zero     * Problem: $\lim_{x \to \infty} x^{e^{-x}}$.     * Form: $\infty^0$ because $e^{-x} \to 0$ as $x \to \infty$.     * Setup: Rewrite as $e^{\lim_{x \to \infty} [e^{-x} \ln x]}$.     * Exponent Limit: $\lim_{x \to \infty} (e^{-x} \ln x)$ is a $0 \times \infty$ product.         * Transform to quotient: $\lim_{x \to \infty} \frac{\ln x}{e^x}$.         * Apply L'Hôpital: $\lim_{x \to \infty} \frac{1/x}{e^x} = \lim_{x \to \infty} \frac{1}{x e^x} = 0$.     * Final Result: $e^0 = 1$.

Indeterminate Differences (Type 4)

• Definition: Occurs during the subtraction of two functions that both approach infinity ($\infty - \infty$).

• Determinant Comparison: Note that $\infty + \infty = \infty$ and $\infty \times \infty = \infty$. These are NOT indeterminate.

• Handling Techniques: There is no universal formula or direct method for $\infty - \infty$. Each problem requires specific algebraic manipulation to convert the difference into a quotient ($\frac{0}{0}$ or $\frac{\infty}{\infty}$).

• Common Algebraic Tricks:     * Trigonometric Identities: Converting terms like $\sec(x)$ and $\tan(x)$ into $\sin(x)$ and $\cos(x)$ to find a common denominator.     * Common Denominator: When dealing with fractions, combining them into a single fraction often reveals a $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form.

• Example 1: Trigonometric Difference     * Problem: $\lim_{x \to \frac{\pi}{2}^-} (\sec x - \tan x)$.     * Analysis: $\sec(\frac{\pi}{2})$ is $\frac{1}{0}$ (infinity) and $\tan(\frac{\pi}{2})$ is infinity. Form: $\infty - \infty$.     * Transformation: Convert to sines and cosines: $\lim_{x \to \frac{\pi}{2}^-} (\frac{1}{\cos x} - \frac{\sin x}{\cos x}) = \lim_{x \to \frac{\pi}{2}^-} \frac{1 - \sin x}{\cos x}$.     * Method: This is now in $\frac{0}{0}$ form.     * Apply L'Hôpital: Derivative of top is $-\cos x$; derivative of bottom is $-\sin x$.     * Final Result: $\frac{-\cos(\frac{\pi}{2})}{-\sin(\frac{\pi}{2})} = \frac{0}{-1} = 0$.

• Example 2: Fractional Difference     * Problem: $\lim_{x \to 1} (\frac{x}{x-1} - \frac{1}{\ln x})$.     * Analysis: As $x \to 1$, denominators approach zero, resulting in $\infty - \infty$.     * Transformation: Find the common denominator to combine into one fraction: $\frac{x \ln x - (x - 1)}{(x - 1) \ln x}$.     * Form: This results in $\frac{0}{0}$ at $x = 1$.     * Derivatives:         * Numerator derivative: Use product rule on $x \ln x$. $\frac{d}{dx}[x \ln x - x + 1] = (1 \times \ln x + x \times \frac{1}{x}) - 1 = \ln x + 1 - 1 = \ln x$.         * Denominator derivative: Use product rule on $(x - 1) \ln x$. $\frac{d}{dx}[(x - 1) \ln x] = 1 \times \ln x + (x - 1) \times \frac{1}{x} = \ln x + \frac{x - 1}{x}$.     * Simplified Expression: $\lim_{x \to 1} \frac{\ln x}{\ln x + \frac{x-1}{x}}$. This is still $\frac{0}{0}$.     * Second L'Hôpital Application:         * Numerator becomes $\frac{1}{x}$.         * Denominator becomes $\frac{1}{x} + \frac{1}{x^2}$. (Note: Derivative of $\frac{x-1}{x} = 1 - \frac{1}{x}$ is $\frac{1}{x^2}$).     * Evaluating at x = 1: $\frac{1/1}{1/1 + 1/1^2} = \frac{1}{1 + 1} = \frac{1}{2}$.

Admin and Preparation

• Scope of Material: The lecture covers sections 6.6 and 6.8 of the curriculum.

• Quiz Information:     * The upcoming quiz covers up to section 7.4.     * Relevant practice problems for the current topic are found in sections 6.6 and 6.8 (specifically mentioned problems 1 through 7).     * Students are advised that questions will be very similar to the examples provided in class.     * Preparation time remaining: approximately 14 hours.