What Could 0^0 Possibly Be? Two Limit Paths, Two Different Answers

Most arithmetic expressions have a clean answer. 2 + 3 = 5. \tfrac{10}{4} = 2.5. You compute, and the answer sits there waiting. But 0^0 is different. Depending on which path you take towards x^y = 0^0, you get a different answer — sometimes 1, sometimes 0, sometimes nothing sensible at all. That is why 0^0 is usually left undefined in school arithmetic. This visualisation shows why.

The slider: two paths towards 0^0

Two curves both heading toward the point where the base and exponent are both zero. Blue (path A): $t^t$, where both base and exponent shrink together. As $t \to 0^+$, the value goes to $1$. Orange (path B): $0^t$, where the base is already zero and the exponent shrinks. For every $t > 0$, the value is exactly $0$. Two paths, two different limiting values — which is why $0^0$ has no single answer.

Path A: the base and exponent shrink together

On this path, the base t and the exponent t move toward zero together. You can compute:

t = 0.5: 0.5^{0.5} = \sqrt{0.5} \approx 0.707
t = 0.1: 0.1^{0.1} \approx 0.794
t = 0.01: 0.01^{0.01} \approx 0.955
t = 0.001: 0.001^{0.001} \approx 0.993
t = 0.0001: 0.0001^{0.0001} \approx 0.9991

The value creeps toward 1 as t shrinks. This makes sense if you think about it: a number just above zero, raised to a very small positive power, is close to 1. The exponent is pulling the answer toward 1 faster than the base is pulling it toward 0. Why: t^t = e^{t \ln t}, and t \ln t \to 0 as t \to 0^+, so e^{t \ln t} \to e^0 = 1.

So along this path, 0^0 = 1 seems correct.

Path B: base is zero from the start

Now consider 0^t. For any positive t, the value is 0^t = 0. Raising zero to a positive power is still zero — you are multiplying a bunch of zeros together, and zero times anything is zero.

t = 0.5: 0^{0.5} = \sqrt{0} = 0
t = 0.1: 0^{0.1} = 0
t = 0.0001: 0^{0.0001} = 0

Along this path, the value is 0 for every positive t. So the limit as t \to 0^+ is also 0. Along this path, 0^0 = 0 seems correct.

The conflict

You have two perfectly sensible-looking arguments giving two different answers:

"Take t \to 0 in t^t." Limit is 1.
"Take t \to 0 in 0^t." Limit is 0.

Both paths arrive at the point where the base and the exponent are both zero, but they disagree on the answer. The expression 0^0 is therefore called an indeterminate form — the value depends on how the 0s got there, not just on the fact that they are both 0. This is the same reason that \tfrac{0}{0}, \infty - \infty, and 1^\infty are indeterminate. Raw arithmetic is not enough; you need the surrounding context.

So is 0^0 = 1 or undefined?

Different parts of mathematics make different choices, and each is defensible inside its own context.

School arithmetic (Class 9 NCERT and up, and JEE Main): 0^0 is usually left undefined. This is the safest choice because it avoids giving students a tool whose rules seem simple but actually depend on context.
Combinatorics and algebra: 0^0 = 1 is the standard convention. The reason is that the formula \sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k} (the binomial theorem) and the formula e^x = \sum_{n=0}^{\infty} \tfrac{x^n}{n!} both require the n = 0, k = 0 term to be 1 when x = y = 0. Using 0^0 = 1 keeps these formulas clean.
Real analysis / calculus: 0^0 is left indeterminate, exactly because of the path-dependence shown above. When you meet 0^0 inside a limit, you use L'Hôpital's rule or a substitution like y = t^t \Rightarrow \ln y = t \ln t to resolve it.

For school-level arithmetic — the world you are in right now — the correct answer is "0^0 is not defined." If an exam question hands you 0^0, that question is almost certainly wrong or a typo.

The related expression a^0 for a \ne 0

It is important not to let the 0^0 weirdness contaminate the clean case. For any a \ne 0, the rule a^0 = 1 is forced by the quotient law and is not debatable. The difficulty is only when the base is exactly zero. See Exponents and Powers for the quotient-law argument, and the neighbouring Why a^0 = 1 Not 0 for the common mistake.

The broader lesson

The 0^0 story is your first real encounter with indeterminate forms, and they come up again everywhere: in the definition of the derivative (\tfrac{0}{0}), in improper integrals (\infty \cdot 0), in series convergence (1^\infty). Whenever arithmetic hands you an ambiguous-looking expression, the fix is to go back to the context — the specific limit path — and evaluate carefully.

This is the first line of a long and interesting story. 0^0 is not a broken symbol; it is a signpost telling you that you have left plain arithmetic and entered limits.