AP® Calculus BC · Unit 1 · Topic 1.7

Selecting Procedures for Determining Limits

Finding a limit is a diagnostic process: try substitution first, and if it fails with 0/0, pick the tool—factoring, a conjugate, a common denominator, or a special trig identity—that matches the algebraic obstruction in front of you.

No single algebra move cracks every limit, because the obstruction blocking direct substitution changes from problem to problem. Sometimes it's a factorable polynomial, sometimes a radical, sometimes a messy fraction inside a fraction, sometimes a trig expression that hides behind sin(x)/x. So limit-finding works like a decision tree: try plugging in first. If that gives a real number, you're done. If it gives 0/0, the function has a removable gap, and you look at its shape to decide the repair. A difference of squares or a common root signals factoring. A square root being added or subtracted signals multiplying by a conjugate. A fraction stacked over another fraction signals clearing denominators first. An expression built from sin(x) or (1 - cos(x)) over x signals the special trig limits, sin(x)/x → 1 and (1 - cos(x))/x → 0 as x → 0.

Each tool does the same job: rewrite the function so the troublesome factor cancels, revealing what the function approaches even though it never actually equals that value at the point itself. This is the heart of why limits exist — they describe behavior near a point, independent of whatever happens (or fails to happen) exactly at that point.

This topic sits right before continuity, where the question becomes whether the limit actually matches the function's value at that point. Selecting procedures well here also sets up derivatives, since every derivative is a limit of a difference quotient that starts life looking like 0/0.

Key terms

Indeterminate form
An expression like 0/0 that gives no information about a limit's value until simplified further.
Removable discontinuity
A point where a function has a gap but the limit still exists because the gap can be filled algebraically.
Conjugate multiplication
Multiplying numerator and denominator by the conjugate of a radical expression to eliminate the square root causing 0/0.
Special trig limits
The facts that sin(x)/x → 1 and (1 - cos x)/x → 0 as x approaches 0, used to resolve trig-based 0/0 forms.

Watch it explained

  1. How it worksHow do I choose between substitution, factoring, conjugates, and the special trig limits?
  2. Worked exampleFind the limit as x approaches 0 of (1 - cos(x))/x2 and name the tool that cracked it.
  3. Worked exampleFind the limit as x approaches 2 of (1/x - 1/2)/(x - 2) by clearing the complex fraction.
  4. Common mistakeWhy doesn't every 0/0 limit come out equal to 1?

Check yourself

Your turn1 of 2
For x not equal to 4, write (1/(x + 4) - 1/(4 + 4))/(x - 4) in simplified form, as a function of x.

End of topic.