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Why must you cancel hh before taking the limit in the difference quotient?

You must cancel hh because direct substitution of h=0h=0 into the difference quotient f(x+h)−f(x)h\frac{f(x+h)-f(x)}{h} results in the indeterminate form 0/00/0. Algebraic simplification, such as expansion and cancellation for polynomials or rationalization for radicals, removes the singularity at h=0h=0, allowing the limit to be evaluated by direct substitution of the simplified expression.

Conditions

  • The function ff is differentiable at xx.
  • The difference quotient initially yields a 0/00/0 form when h=0h=0.
  • Algebraic manipulation is valid for nonzero hh.

Reasoning, step by step

  1. Identify that substituting h=0h=0 directly gives 0/00/0.
  2. Perform algebraic simplification (expansion, factoring, or conjugate multiplication).
  3. Cancel the common factor hh from the numerator and denominator.
  4. Evaluate the limit of the simplified expression as h→0h \to 0.

Example

The script demonstrates this for f(x)=x2f(x)=x^2: 'For nonzero h the quotient becomes 2x+h2x+h, whose limit is 2x.' Without canceling hh, the expression remains undefined at h=0h=0.

Common misconceptions

  • Believing that 0/00/0 equals zero or one.
  • Thinking that the limit can be taken before simplification.
  • Assuming that cancellation is valid at h=0h=0 itself, rather than for h≠0h \neq 0 approaching 0.

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