AP® Calculus BC · Unit 1 · Topic 1.6

Determining Limits Using Algebraic Manipulation

When direct substitution gives 0/0, the function has a removable hole, not a nonexistent limit. Factoring or multiplying by a conjugate cancels the troublesome factor so you can substitute again.

Plugging in a number and getting 0/0 doesn't mean the limit fails to exist. It means the expression is hiding a common factor that vanishes at that exact x-value, creating a hole in the graph rather than a jump or asymptote. Your job is to rewrite the expression so that factor is gone, then substitute again.

For rational expressions, factor the numerator and denominator and cancel whatever matches. For expressions with square roots, multiply top and bottom by the conjugate of the radical part, which uses the difference of squares to turn a square root into a plain variable that can cancel with the denominator. Either way, the algebra doesn't change what the function approaches, it just removes the one point where substitution was blocked.

This is the main algebraic toolkit before limits get harder: it sets up the idea that a limit only cares about behavior near a point, not at it, which is exactly what makes continuity meaningful later. When factoring and conjugates don't resolve the indeterminate form, later topics bring in more powerful tools like L'Hopital's rule, but for now, these two moves handle the overwhelming majority of 0/0 limits you'll see.

Key terms

Indeterminate form
An expression like 0/0 that gives no information on its own and signals more algebra is needed.
Removable discontinuity
A hole in a graph at a point where the limit exists but the function is undefined or disagrees with that limit.
Conjugate
The expression formed by flipping the sign between two terms, used to rationalize square roots via difference of squares.
Direct substitution
Plugging the approached x-value straight into the function to find a limit, valid only when it gives a defined result.

Watch it explained

  1. How it worksHow does factoring and cancelling erase the hole that was blocking substitution?
  2. Worked exampleFind the limit as x approaches 4 of (x2 - 16)/(x - 4).
  3. Worked exampleFind the limit as x approaches 0 of (√(x + 9) - 3)/x using the conjugate.
  4. Common mistakeWhy isn't cancelling a factor of (x - 4) quietly turning this into a different function?

Check yourself

Your turn1 of 2
For x not equal to 0, write (√(x + 12) - √12)/x in simplified form (after rationalizing with the conjugate), as a function of x.

End of topic.