The forms $0^0$, $\infty^0$, and $1^\infty$ are all indeterminate. Why? Consider:
The answer depends on the specific functions involved. To find out, we use a powerful technique: take the logarithm.
| Property | Value |
|---|---|
| Section | Stewart 6.8 |
| Course | MATH162 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
| Form | Example | Why Indeterminate |
|---|---|---|
| $0^0$ | $\lim_{x \to 0^+} x^x$ | Base shrinks, exponent shrinks |
| $\infty^0$ | $\lim_{x \to \infty} x^{1/x}$ | Base grows, exponent shrinks |
| $1^\infty$ | $\lim_{x \to \infty} (1 + 1/x)^x$ | Base approaches 1, exponent grows |
Key insight: For $y = f(x)^{g(x)}$, take the natural log: $$\ln y = g(x) \cdot \ln[f(x)]$$
This converts the power into a product, which we know how to handle!
Step 1: Let $y = [f(x)]^{g(x)}$ where the limit is $0^0$, $\infty^0$, or $1^\infty$.
Step 2: Take the natural log: $\ln y = g(x) \cdot \ln[f(x)]$.
Step 3: Evaluate $\lim_{x \to a} \ln y$ (this is usually $0 \cdot \infty$; convert and use L'Hospital's).
Step 4: If $\lim \ln y = L$, then $\lim y = e^L$.
$$\boxed{\lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} g(x) \ln f(x)}}$$
The exponential function is continuous, so: $$\lim y = \lim e^{\ln y} = e^{\lim \ln y}$$
We can "pull the limit inside" the exponential.
Problem: Evaluate $\lim_{x \to 0^+} x^x$.
Step 1: Identify the form.
Step 2: Let $y = x^x$. Take the log: $$\ln y = x \ln x$$
Step 3: Evaluate $\lim_{x \to 0^+} x \ln x$.
This is $0 \cdot (-\infty)$. Convert: $$x \ln x = \frac{\ln x}{1/x}$$
Form: $\frac{-\infty}{\infty}$. Apply L'Hospital's: $$\lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0$$
Step 4: Therefore: $$\lim_{x \to 0^+} x^x = e^0 = 1$$
Answer: $\lim_{x \to 0^+} x^x = 1$
Problem: Evaluate $\lim_{x \to \infty} x^{1/x}$.
Step 1: Identify the form.
Step 2: Let $y = x^{1/x}$. Take the log: $$\ln y = \frac{1}{x} \cdot \ln x = \frac{\ln x}{x}$$
Step 3: Evaluate $\lim_{x \to \infty} \frac{\ln x}{x}$.
Form: $\frac{\infty}{\infty}$. Apply L'Hospital's: $$\lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{1/x}{1} = 0$$
Step 4: Therefore: $$\lim_{x \to \infty} x^{1/x} = e^0 = 1$$
Answer: $\lim_{x \to \infty} x^{1/x} = 1$
Problem: Evaluate $\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x$.
Step 1: Identify the form.
Step 2: Let $y = (1 + 1/x)^x$. Take the log: $$\ln y = x \cdot \ln\left(1 + \frac{1}{x}\right)$$
Step 3: Evaluate the limit.
Form: $\infty \cdot 0$. Convert to quotient: $$\ln y = \frac{\ln(1 + 1/x)}{1/x}$$
Form: $\frac{0}{0}$. Apply L'Hospital's: $$\lim_{x \to \infty} \frac{\ln(1 + 1/x)}{1/x}$$
Derivatives:
$$= \lim_{x \to \infty} \frac{-1/(x^2 + x)}{-1/x^2} = \lim_{x \to \infty} \frac{x^2}{x^2 + x} = \lim_{x \to \infty} \frac{1}{1 + 1/x} = 1$$
Step 4: Therefore: $$\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e^1 = e$$
Answer: $\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e$
This is the definition of $e$!
| Form | Example | $\lim \ln y$ | Answer |
|---|---|---|---|
| $0^0$ | $x^x$ as $x \to 0^+$ | $0$ | $e^0 = 1$ |
| $\infty^0$ | $x^{1/x}$ as $x \to \infty$ | $0$ | $e^0 = 1$ |
| $1^\infty$ | $(1+1/x)^x$ as $x \to \infty$ | $1$ | $e^1 = e$ |
For $1^\infty$ forms: The answer is often $e^k$ for some constant $k$.
If $\lim_{x \to a}(1 + f(x))^{g(x)}$ where $f(x) \to 0$ and $g(x) \to \infty$: $$= e^{\lim_{x \to a} f(x) \cdot g(x)}$$
Example: $\lim_{x \to \infty}(1 + 3/x)^{2x} = e^{\lim 3/x \cdot 2x} = e^6$
Evaluate $\lim_{x \to 0^+} x^{2x}$.
Evaluate $\lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{2x}$.
Evaluate $\lim_{x \to 0^+} (\cos x)^{1/x^2}$.
Evaluate $\lim_{x \to 0^+} (1 + 2\sin x)^{\csc 3x}$.
Prove that if $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = \infty$, and $\lim_{x \to a} f(x) \cdot g(x) = L$ exists, then:
$$\lim_{x \to a} [1 + f(x)]^{g(x)} = e^L$$
Use this to find $\lim_{x \to 0} (1 + x^2)^{1/x^2}$.
Question 1: Why is $0^\infty$ NOT an indeterminate form?
When the base approaches $0^+$ (positive and small) and the exponent approaches $+\infty$:
$$\lim_{x \to a} [f(x)]^{g(x)} = 0$$
Think of $(0.1)^{100}$, $(0.01)^{1000}$, etc. A small positive base raised to a huge power gives an even smaller number.
This is NOT indeterminate because the answer is always 0 regardless of the specific functions.
Question 2: A student says "$1^\infty = 1$ because $1$ to any power is $1$." What's the error?
The error is treating the limit as static arithmetic. When we write $1^\infty$, we mean:
$$\lim_{x \to a} [f(x)]^{g(x)}$$ where $f(x) \to 1$ and $g(x) \to \infty$
The base isn't exactly 1. It is approaching 1. A number slightly different from 1, raised to a huge power, can give any result.
Example: $(1 + 1/n)^n \to e \approx 2.718$, not 1!
The slight deviation from 1, compounded infinitely many times, creates a finite but nontrivial limit.
The Logarithm Bridge: Indeterminate powers are stuck on one side of a river. The logarithm is a bridge that takes them across to "Product Land," where we know how to work. After solving the problem in Product Land, we take the exponential bridge back.
$$[f]^g \xrightarrow{\ln} g \cdot \ln f \xrightarrow{\text{L'Hop}} L \xrightarrow{e^{(\cdot)}} e^L$$
Looking back:
Looking ahead:
| Previous | Up | Next |
|---|---|---|
| Converting Indeterminate Products | Skills Index | Improper Integrals |
Last updated: 2026-01-22