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How does clearing the inner fraction resolve the indeterminate limit of (3+3x3+3x)/(1/x+11/x+1) as x approaches -1?

Clearing the inner fraction resolves the limit by simplifying the complex rational expression into a polynomial form. Near x=−1x=-1, with x≠−1x\ne-1, multiplying the numerator and denominator of 3+3x1/x+1\frac{3+3x}{1/x+1} by xx gives 3x+3x21+x\frac{3x+3x^2}{1+x}. Factoring the numerator yields 3x(1+x)1+x\frac{3x(1+x)}{1+x}, which simplifies to 3x3x after cancelling (1+x)(1+x). Thus, the limit as x→−1x \to -1 is −3-3.

Conditions

  • The limit is as x→−1x \to -1.
  • The expression contains an inner fraction 1/x1/x in the denominator.
  • Multiplication by xx is valid for x≠0x \ne 0, and specifically near x=−1x=-1 where x≠0x \ne 0.
  • The original expression is undefined at x=−1x=-1 (indeterminate form 0/00/0).

Reasoning, step by step

  1. Identify the indeterminate form 0/00/0 by substituting x=−1x=-1.
  2. Multiply the numerator and denominator by xx to clear the inner fraction 1/x1/x.
  3. Simplify the new expression: x(3+3x)x(1/x+1)=3x+3x21+x\frac{x(3+3x)}{x(1/x+1)} = \frac{3x+3x^2}{1+x}.
  4. Factor the numerator: 3x(1+x)3x(1+x).
  5. Cancel the common factor (1+x)(1+x), valid for x≠−1x \ne -1.
  6. Evaluate the limit of the simplified expression 3x3x as x→−1x \to -1.
  7. Calculate the final value: 3(−1)=−33(-1) = -3.

Example

The script states: "Near x=−1x=-1, with x≠−1x\ne-1, multiplying the numerator and denominator of 3+3x1/x+1\frac{3+3x}{1/x+1} by xx gives 3x+3x21+x=3x\frac{3x+3x^2}{1+x}=3x. Thus the limit is −3-3."

Common misconceptions

  • Multiplying only the denominator by xx and forgetting the numerator.
  • Believing that a zero denominator alone is a general license to cancel without checking the indeterminate form.
  • Confusing the point x=0x=0 (where the inner fraction is undefined) with the limit point x=−1x=-1.

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