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How does multiplying by the conjugate resolve the indeterminate limit of (x-5)/(sqrt(x+4x+4)-3) as x approaches 5?

Multiplying by the conjugate resolves the limit by eliminating the radical in the denominator, which causes the 0/00/0 indeterminacy. For lim⁡x→5x−5x+4−3\lim_{x\to5}\frac{x-5}{\sqrt{x+4}-3}, multiply the numerator and denominator by x+4+3x+4+3\frac{\sqrt{x+4}+3}{\sqrt{x+4}+3}. The difference of squares in the denominator produces x−5x-5, which cancels with the numerator. The remaining expression approaches 3+3=63+3=6.

Conditions

  • The limit is as x→5x \to 5.
  • The expression involves a radical in the denominator causing a 0/00/0 form.
  • The conjugate is x+4+3\sqrt{x+4}+3.

Reasoning, step by step

  1. Identify the indeterminate form 0/00/0 by substituting x=5x=5.
  2. Multiply the numerator and denominator by the conjugate of the denominator: x+4+3\sqrt{x+4}+3.
  3. Apply the difference of squares formula (a−b)(a+b)=a2−b2(a-b)(a+b)=a^2-b^2 to the denominator: (x+4)2−32=x+4−9=x−5(\sqrt{x+4})^2 - 3^2 = x+4-9 = x-5.
  4. Cancel the common factor (x−5)(x-5) from the numerator and denominator.
  5. Evaluate the limit of the remaining expression x+4+3\sqrt{x+4}+3 as x→5x \to 5.
  6. Calculate the final value: 5+4+3=3+3=6\sqrt{5+4}+3 = 3+3=6.

Example

The script states: "For lim⁡x→5x−5x+4−3\lim_{x\to5}\frac{x-5}{\sqrt{x+4}-3}, multiply by x+4+3x+4+3\frac{\sqrt{x+4}+3}{\sqrt{x+4}+3}. The difference of squares produces x−5x-5 in the denominator; after cancellation the remaining expression approaches 3+3=63+3=6."

Common misconceptions

  • Forgetting to multiply the numerator by the conjugate as well.
  • Incorrectly expanding the denominator instead of using the difference of squares.
  • Believing that the conjugate method works for all types of indeterminate limits, not just those involving radicals.

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