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When does the limit definition of the derivative fail to produce a finite value for f(x)=xf(x)=\sqrt{x}?

The limit definition fails to produce a finite derivative for f(x)=xf(x)=\sqrt{x} at x=0x=0. While the formula yields 12x\frac{1}{2\sqrt{x}} for x>0x>0, substituting x=0x=0 results in division by zero, indicating that the slope of the tangent line is vertical and the derivative is undefined (infinite) at that point.

Conditions

  • f(x)=xf(x)=\sqrt{x}.
  • The point of evaluation is x=0x=0.
  • The limit is taken as h→0+h \to 0^+ (since the domain is x≥0x \ge 0).

Reasoning, step by step

  1. Apply the derivative formula derived for x>0x>0: f′(x)=12xf'(x) = \frac{1}{2\sqrt{x}}.
  2. Attempt to evaluate at x=0x=0.
  3. Observe that the denominator becomes zero.
  4. Conclude that the limit is infinite, so the finite derivative does not exist.

Example

The script states: 'At x>0x>0 the limit yields 1/(2x)1/(2\sqrt{x}); the formula does not give a finite derivative at x=0x=0.'

Common misconceptions

  • Assuming the derivative exists everywhere the function is defined.
  • Confusing the function's continuity at x=0x=0 with its differentiability.
  • Believing that a vertical tangent implies a derivative of zero.

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