L'Hospital's Rule for Indeterminate Powers

5 min read

When the Exponent Matters Most

What is $\lim_{x \to 0^+} x^x$? If you think “zero raised to any power is zero,” you get 0. If you think “anything raised to the zero power is 1,” you get 1. Both intuitions fail because this is $0^0$, an indeterminate power form.

The three indeterminate power forms are $0^0$, $\infty^0$, and $1^\infty$. They all share a common solution strategy: take the logarithm to convert the power into a product, then apply the techniques from the previous skill.

This technique is essential for understanding limits like $\lim_{x \to 0^+}(1 + x)^{1/x} = e$, the foundation of continuous compounding.

Prerequisite Map

Quick Reference

Key Concepts

The Three Indeterminate Power Forms

The Logarithm Strategy

For $\lim_{x \to a} [f(x)]^{g(x)}$ in an indeterminate power form:

Method 1: Let $y = [f(x)]^{g(x)}$, then take ln

$$\ln y = \ln\left([f(x)]^{g(x)}\right) = g(x) \ln f(x)$$

Find $\lim_{x \to a} \ln y$, then recover $y = e^{\ln y}$.

Method 2: Rewrite as exponential directly

$$[f(x)]^{g(x)} = e^{g(x) \ln f(x)}$$

Then find $\lim_{x \to a} g(x) \ln f(x)$.

Both Methods Lead to the Same Place

$$\boxed{\lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} g(x) \ln f(x)}}$$

The key insight: indeterminate powers become indeterminate products after taking the logarithm, and we know how to handle those!

Form Analysis

All three reduce to the same type of indeterminate product!

The Algorithm

INDETERMINATE POWER: [f(x)]^g(x)
──────────────────────────────────────
              ↓
    Let y = [f(x)]^g(x)
              ↓
    Take ln: ln y = g(x) · ln f(x)
              ↓
    This is form 0 · ∞ or ∞ · 0
              ↓
    Convert to quotient:
    ln y = ln f(x) / (1/g(x))  OR
    ln y = g(x) / (1/ln f(x))
              ↓
    Apply L'Hospital's Rule
              ↓
    Find L = lim(ln y)
              ↓
    Answer: lim y = e^L

Worked Examples

Example 1: Form $0^0$

Evaluate $\lim_{x \to 0^+} x^x$

Step 1: Identify form: $0^0$ (indeterminate power) ✓

Step 2: Take logarithm $$y = x^x \implies \ln y = x \ln x$$

Step 3: This is form $0 \cdot (-\infty)$. Convert: $$\ln y = \frac{\ln x}{1/x} \quad \text{(form } \frac{-\infty}{\infty}\text{)}$$

Step 4: Apply L’Hospital’s Rule: $$\lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0$$

Step 5: Recover original limit: $$\lim_{x \to 0^+} y = e^{\lim \ln y} = e^0 = 1$$

Answer: $\lim_{x \to 0^+} x^x = 1$

Example 2: Form $\infty^0$

Evaluate $\lim_{x \to \infty} x^{1/x}$

Step 1: Form is $\infty^0$ ✓

Step 2: Take logarithm $$y = x^{1/x} \implies \ln y = \frac{\ln x}{x}$$

Step 3: This is already a quotient! Form $\frac{\infty}{\infty}$ ✓

Step 4: Apply L’Hospital’s Rule: $$\lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{1/x}{1} = 0$$

Step 5: Recover: $$\lim_{x \to \infty} x^{1/x} = e^0 = 1$$

Answer: $\lim_{x \to \infty} x^{1/x} = 1$

Example 3: Form $1^\infty$ (The Most Important!)

Evaluate $\lim_{x \to 0^+} (1 + x)^{1/x}$

Step 1: Form is $1^\infty$ ✓ (base → 1, exponent → ∞)

Step 2: Take logarithm $$y = (1+x)^{1/x} \implies \ln y = \frac{\ln(1+x)}{x}$$

Step 3: Form $\frac{0}{0}$ ✓

Step 4: Apply L’Hospital’s Rule: $$\lim_{x \to 0^+} \frac{\ln(1+x)}{x} = \lim_{x \to 0^+} \frac{1/(1+x)}{1} = \frac{1}{1} = 1$$

Step 5: Recover: $$\lim_{x \to 0^+} (1+x)^{1/x} = e^1 = e$$

Answer: $\lim_{x \to 0^+} (1+x)^{1/x} = e$

This is one of the most important limits in calculus: it defines $e$.

Example 4: A Harder $1^\infty$ Form

Evaluate $\lim_{x \to 0^+} (1 + 2\sin x)^{\csc 3x}$

Step 1: As $x \to 0^+$: base $1 + 2\sin 0 = 1$, exponent $\csc 3x = 1/\sin 3x \to +\infty$. Form: $1^\infty$ ✓

Step 2: Take logarithm $$\ln y = \csc 3x \cdot \ln(1 + 2\sin x) = \frac{\ln(1 + 2\sin x)}{\sin 3x}$$

Step 3: Form $\frac{0}{0}$ ✓

Step 4: Apply L’Hospital’s Rule (chain rule on both!): $$\lim_{x \to 0^+} \frac{\ln(1 + 2\sin x)}{\sin 3x} = \lim_{x \to 0^+} \frac{\frac{2\cos x}{1 + 2\sin x}}{3\cos 3x}$$

Step 5: Evaluate at $x = 0$: $$= \frac{\frac{2 \cdot 1}{1 + 0}}{3 \cdot 1} = \frac{2}{3}$$

Step 6: Recover: $$\lim_{x \to 0^+} (1 + 2\sin x)^{\csc 3x} = e^{2/3}$$

Practice Problems

Common Misconceptions

Mastery Checklist

• I can identify $0^0$, $\infty^0$, and $1^\infty$ as indeterminate power forms
• I know to take the logarithm to convert powers to products
• I can evaluate $\lim_{x \to 0^+} x^x = 1$
• I can evaluate $\lim_{x \to \infty} x^{1/x} = 1$
• I know that $\lim_{x \to 0^+}(1+x)^{1/x} = e$
• I can apply the pattern $(1 + k/x)^x \to e^k$
• I understand the connection to continuous compounding

Mental Model

The “Exponential Detour”:

When you see a power with a variable base AND variable exponent, think: “I can’t directly evaluate this, but I can take a detour through exponentials.”

$$[f]^g = e^{g \ln f}$$

Now instead of dealing with a power, you’re finding the exponent of $e$. The limit $g \ln f$ is a product, which you already know how to handle.

The Pattern to Remember:

$$\text{Indeterminate power} \xrightarrow{\ln} \text{Indeterminate product} \xrightarrow{\text{convert}} \text{Quotient} \xrightarrow{\text{L'H}} \text{Number} \xrightarrow{e^{(\cdot)}} \text{Answer}$$

Connections

Looking back:

• L’Hospital’s Rule (Products & Differences): powers reduce to products
• Logarithm Properties: essential for $\ln(f^g) = g \ln f$

Looking ahead:

• Continuous Compounding: the $1^\infty$ form explains why $A = A_0 e^{rt}$
• Taylor Series: another approach to these limits

The Big Picture:

• $\lim_{n \to \infty}(1 + 1/n)^n = e$ is one definition of $e$
• This connects compound interest, exponential growth, and the natural logarithm

Last updated: 2026-01-22