Why does the epsilon-delta definition require the condition to hold for every epsilon?

Conditions

• $\epsilon$ represents an arbitrary positive tolerance.
• The definition aims to prove convergence to a specific limit $L$.

Reasoning, step by step

1. Recognize that a single $\epsilon$ only provides a finite bound on the error.
2. Understand that 'limit' implies the error can be reduced without bound.
3. Require that for *any* chosen $\epsilon$ (no matter how small), a corresponding $\delta$ exists.
4. Conclude that this universal requirement forces $f(x)$ to approach $L$ infinitely closely.

Example

The speaker warns: 'that was just a specific example... But in order for this... by definition... it doesn't just work for one specific instance, it works for any number you give me.' He mentions even extremely small numbers like $10^{-100}$.

Common misconceptions

• Believing that finding a delta for one nice epsilon (like 0.5) proves the limit.
• Thinking that the definition allows for a fixed maximum error.
• Confusing the existence of a bound with the convergence to a specific value.

Watch the explanation

Explore next

• How does the choice of a very small epsilon affect the size of delta?
• What is the logical structure of the 'for all epsilon, there exists delta' statement?
• Can a function satisfy the condition for one epsilon but fail for another?