Navigation: Wiki Home > Skills > L'Hospital's Rule for Indeterminate Powers
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.
| Property | Value |
|---|---|
| Concept | Indeterminate Forms & L'Hospital's Rule |
| Chapter | Chapter 6, Section 8 |
| Difficulty | Advanced |
| Time | ~20 minutes |
| Form | Base → | Exponent → | Example |
|---|---|---|---|
| $0^0$ | $0^+$ | $0$ | $\lim_{x \to 0^+} x^x$ |
| $\infty^0$ | $+\infty$ | $0$ | $\lim_{x \to \infty} x^{1/x}$ |
| $1^\infty$ | $1$ | $\pm\infty$ | $\lim_{x \to 0^+} (1+x)^{1/x}$ |
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 $\lim_{x \to a} g(x) \ln f(x)$.
$$\boxed{\lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} g(x) \ln f(x)}}$$
The key insight: indeterminate powers become indeterminate products after taking the logarithm, and we know how to handle those!
| Original Form | After $\ln$ | Product Type |
|---|---|---|
| $0^0$ | $0 \cdot (-\infty)$ | $0 \cdot \infty$ |
| $\infty^0$ | $0 \cdot \infty$ | $0 \cdot \infty$ |
| $1^\infty$ | $\pm\infty \cdot 0$ | $0 \cdot \infty$ |
All three reduce to the same type of indeterminate product!
INDETERMINATE POWER: [f(x)]^g(x)
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Let y = [f(x)]^g(x)
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Take ln: ln y = g(x) · ln f(x)
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This is form 0 · ∞ or ∞ · 0
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Convert to quotient:
ln y = ln f(x) / (1/g(x)) OR
ln y = g(x) / (1/ln f(x))
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Apply L'Hospital's Rule
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Find L = lim(ln y)
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Answer: lim y = e^LEvaluate $\lim_{x \to 0^+} x^x$
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: $\lim_{x \to 0^+} x^x = 1$
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: $\lim_{x \to \infty} x^{1/x} = 1$
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 $e$!
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: $$\lim_{x \to 0^+} (1 + 2\sin x)^{\csc 3x} = e^{2/3}$$
Classify each limit as $0^0$, $\infty^0$, $1^\infty$, or NOT an indeterminate power form:
(a) $\lim_{x \to 0^+} x^{\sin x}$
(b) $\lim_{x \to \infty} (1 + 1/x)^x$
(c) $\lim_{x \to 0^+} (\cos x)^{1/x^2}$
(d) $\lim_{x \to \infty} 2^{1/x}$
Evaluate $\lim_{x \to 0^+} x^{x^2}$
Evaluate $\lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{2x}$
The value of an investment of $A_0$ dollars at annual interest rate $r$, compounded $n$ times per year for $t$ years, is:
$$A = A_0\left(1 + \frac{r}{n}\right)^{nt}$$
(a) Show that as $n \to \infty$ (continuous compounding), $A \to A_0 e^{rt}$.
(b) If you invest $1000 at 5% annual interest for 10 years, what is the difference between annual compounding ($n=1$) and continuous compounding?
Prove that for any constant $k$:
$$\lim_{x \to \infty} \left(1 + \frac{k}{x}\right)^x = e^k$$
Then use this to explain why:
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:
$$\text{Indeterminate power} \xrightarrow{\ln} \text{Indeterminate product} \xrightarrow{\text{convert}} \text{Quotient} \xrightarrow{\text{L'H}} \text{Number} \xrightarrow{e^{(\cdot)}} \text{Answer}$$
Looking back:
Looking ahead:
The Big Picture:
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|---|---|---|
| L'Hospital's Rule (Products & Differences) | Skills Index | Exponential Growth |
Last updated: 2026-01-22