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How does the ceiling function ensure N is an integer in the epsilon-N definition?

The algebraic solution for the threshold often yields a real number, such as 9ε+3\sqrt{\frac{9}{\varepsilon} + 3}. Since the definition of a sequence limit requires NN to be a positive integer, we apply the ceiling function ⌈x⌉\lceil x \rceil, which rounds up to the nearest integer. This guarantees that the chosen NN is strictly greater than or equal to the real-valued threshold, ensuring all n>Nn > N satisfy the error condition.

Conditions

  • NN must be a positive integer.
  • The algebraic solution yields a real number.
  • Ceiling function ⌈x⌉\lceil x \rceil rounds up to the nearest integer.

Reasoning, step by step

  1. Solve the inequality for nn to get a real-valued bound, e.g., n>4.58n > 4.58.
  2. Apply the ceiling function to this bound: ⌈4.58⌉=5\lceil 4.58 \rceil = 5.
  3. Set N=5N = 5.
  4. Verify that for all integers n>5n > 5, the condition holds.

Example

For ε=0.5\varepsilon = 0.5, the calculation gives N=⌈4.58⌉=5N = \lceil 4.58 \rceil = 5. This ensures NN is an integer.

Common misconceptions

  • Using the floor function instead of the ceiling function, which might result in an NN that is too small.
  • Believing NN can be a non-integer real number.

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