What is the formal epsilon-delta definition of the limit of a function?

Conditions

• $\epsilon$ is an arbitrary positive real number
• $\delta$ is a positive real number dependent on $\epsilon$
• $x$ is in the domain of $f$ and $x \neq a$

Reasoning, step by step

1. Choose an arbitrary tolerance $\epsilon > 0$ for the output distance from $L$.
2. Find a corresponding input radius $\delta > 0$ around $a$.
3. Verify that for all $x$ satisfying $0 < |x - a| < \delta$, the inequality $|f(x) - L| < \epsilon$ holds.
4. Conclude that $\lim_{x \to a} f(x) = L$ if this condition is met for every possible $\epsilon$.

Example

The board explicitly displays the implication: $0<|x-a|<\delta \Rightarrow |f(x)-L|<\epsilon$. The speaker describes this as a game where one person gives an $\epsilon > 0$, and the other must provide a $\delta$ that makes the condition true.

Common misconceptions

• Believing that $\delta$ is chosen first and $\epsilon$ follows; in reality, $\epsilon$ is prescribed first, and $\delta$ is supplied in response.
• Thinking that the condition $|x - a| < \delta$ is sufficient; the definition requires the punctured neighborhood $0 < |x - a| < \delta$ to exclude the point $x = a$ itself.
• Assuming the definition only needs to work for one specific $\epsilon$; it must hold for *every* positive $\epsilon$, no matter how small.

Watch the explanation

Explore next

• Why does the definition use $0 < |x - a|$ instead of just $|x - a| < \delta$?
• How does the epsilon-delta definition relate to the intuitive idea of a limit?
• What is the geometric interpretation of the epsilon band and delta interval?