Rationalizing Indeterminate limits|Rationalization of function|Rational expression|Limit of function

*Lecture – Rationalising Indeterminate Limits (text only)*

Why do we get 0/0 or ∞/∞? When you plug the limit point into the expression you obtain an undefined form, so you must reshape the expression without changing its value.

The rationalisation trick is handy when a square‑root (or higher‑order root) creates the 0/0 situation. Multiply the numerator and denominator by the conjugate of the part that contains the root. This removes the root and lets you cancel the troublesome factor.

*Example:* limit as x approaches 4 of (sqrt(x) – 2) / (x – 4)

Direct substitution gives 0/0, which is indeterminate.

1. Identify the conjugate of the numerator: sqrt(x) + 2.
2. Multiply the whole fraction by (sqrt(x) + 2) / (sqrt(x) + 2).
3. The numerator becomes (sqrt(x))^2 – 2^2 = x – 4.
4. Cancel the common factor (x – 4) from numerator and denominator, leaving 1 / (sqrt(x) + 2).
5. Now substitute x = 4: sqrt(4) = 2, so 1 / (2 + 2) = 1/4.

Therefore the limit equals 1/4.

*When to use rationalisation:* use it when you see a root in the numerator or denominator that becomes zero at the limit point, and the expression is of the 0/0 form after substitution. Multiplying by the conjugate removes the root without changing the limit because you are essentially multiplying by 1.

*Quick practice:* limit as x approaches 9 of (sqrt(x) – 3) / (x – 9). (Answer: 1/6)

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