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How can two admissible sequences with different output limits disprove the existence of a function limit?

If two sequences xnx_n and yny_n both approach x0x_0 (with terms not equal to x0x_0) but their function values f(xn)f(x_n) and f(yn)f(y_n) approach different limits AA and BB (A≠BA \neq B), then the function limit lim⁡x→x0f(x)\lim_{x \to x_0} f(x) does not exist.

Conditions

  • Both sequences must approach the same accumulation point x0x_0.
  • Neither sequence can contain the point x0x_0 itself.
  • The limits of the output sequences must be distinct (A≠BA \neq B).

Reasoning, step by step

  1. Identify two distinct sequences xnx_n and yny_n in the domain approaching x0x_0.
  2. Calculate the limit of f(xn)f(x_n) as n→∞n \to \infty, denoted AA.
  3. Calculate the limit of f(yn)f(y_n) as n→∞n \to \infty, denoted BB.
  4. Verify that A≠BA \neq B.
  5. Conclude that the function limit does not exist because the sequential criterion requires all such sequences to yield the same limit.

Example

For f(x)=sin⁡(1/x)f(x) = \sin(1/x) at x0=0x_0=0, choosing xn=1/(2πn+π/2)x_n = 1/(2\pi n + \pi/2) yields f(xn)=1f(x_n)=1, while yn=1/(2πn+3π/2)y_n = 1/(2\pi n + 3\pi/2) yields f(yn)=−1f(y_n)=-1. Since 1≠−11 \neq -1, the limit does not exist.

Common misconceptions

  • Assuming that checking only one sequence is sufficient to prove non-existence; you must find at least two conflicting sequences.
  • Believing that the sequences must approach from opposite sides; they can approach from the same side as long as the outputs differ.

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