Evaluating Limits by Direct Substitution and Factoring
Simple Explanation
For most "well-behaved" functions, a limit can be found by direct substitution — just plug in x=a. But if this gives an indeterminate form like 0/0, factor both the numerator and denominator first, cancel any common factor, and THEN substitute.
Why Do We Need It?
This is the practical, everyday technique for evaluating most limits by hand, without needing a table of values or a graph.
Worked Example
Evaluate a limit using factoring
Find lim(x→2) (x²−4)/(x−2).
Why Does This Work?
The cancelled expression x+2 agrees with the original (x²−4)/(x−2) at EVERY x-value except exactly x=2 — and since a limit only cares about values near (not at) x=2, evaluating the simplified expression at x=2 correctly gives the limit of the original.
Real-Life Example
A formula with a removable gap at one operating condition
An engineering formula for efficiency happens to produce 0/0 at exactly one specific input setting, even though efficiency clearly should have a well-defined value there.
Factoring and cancelling (just as with a limit) reveals the true, well-defined value the formula approaches at that exact setting.
Practice
Find lim(x→5) (x²−25)/(x−5).
MediumCommon mistake
Giving up and saying "undefined" the moment direct substitution produces 0/0 — this only means try factoring next; the limit itself may still exist.
Quick Review
- Try direct substitution first.
- If you get 0/0, factor numerator and denominator, cancel common factors, then substitute again.
- The limit can exist and be found this way even when the original expression is undefined at that point.