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What distinguishes the behavior of x2x^2 from sin⁡(1/x)\sin (1/x) near zero in the context of limits?

x2x^2 exhibits stable, predictable convergence where all sequences approaching 0 yield outputs approaching 0. In contrast, sin⁡(1/x)\sin(1/x) oscillates infinitely rapidly near 0, allowing the construction of sequences with distinct output limits (e.g., 1 and -1), thus failing the sequential criterion.

Conditions

  • Both functions are analyzed at the accumulation point x0=0x_0 = 0.
  • The comparison focuses on the behavior of f(xn)f(x_n) for sequences xn→0x_n \to 0.
  • xn≠0x_n \neq 0 for all sequences considered.

Reasoning, step by step

  1. Analyze f(x)=x2f(x) = x^2: Show that xn→0  ⟹  xn2→0x_n \to 0 \implies x_n^2 \to 0 universally.
  2. Analyze f(x)=sin⁡(1/x)f(x) = \sin(1/x): Construct specific sequences xnx_n and yny_n approaching 0.
  3. Evaluate f(xn)f(x_n) and f(yn)f(y_n) for sin⁡(1/x)\sin(1/x) to find different limits.
  4. Contrast the universal consistency of x2x^2 with the path-dependence/oscillation of sin⁡(1/x)\sin(1/x).
  5. Conclude that x2x^2 has a limit, while sin⁡(1/x)\sin(1/x) does not.

Example

The animation contrasts the consistent behavior of x2x^2 with the oscillations of sin⁡(1/x)\sin(1/x). For x2x^2, outputs smooth to 0. For sin⁡(1/x)\sin(1/x), outputs jump between 1 and -1.

Common misconceptions

  • Thinking that sin⁡(1/x)\sin(1/x) has a limit of 0 because it is 'average' 0; the limit requires strict convergence, not averaging.
  • Believing that x2x^2's limit depends on the direction of approach; it does not, due to symmetry and continuity.

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