Module 3 · Lesson 3.2

Calculating Limits — Algebraic Techniques & Indeterminate Forms

Learn how to cancel the algebraic “hole” that produces an indeterminate form so a limit can be evaluated by direct substitution.

Intuition

Sometimes plugging the target value straight into a function gives $\frac{0}{0}$. That is an indeterminate form—not “undefined forever,” just a signal that the expression is hiding a removable discontinuity, a hole you can cancel with algebra.

Factor the numerator and denominator, cancel the common factor that is zero at the target, and the remaining simplified expression is continuous there. Its value is the limit. The same cleaning step appears in sensor-error correction: raw readings that look singular become usable once the shared error term is removed.

Core concept

Canceling the hole means rewriting the original expression so that the factor causing both numerator and denominator to vanish is removed. After cancellation the limit equals the value of the simplified function at the target point.

Expression$\dfrac{x^{2}-4}{x-2}$
At $x=2$$\dfrac{0}{0}$ (indeterminate)
Factor$\dfrac{(x-2)(x+2)}{x-2}$
Cancel$x+2$ ($x\neq 2$)
Limit$2+2=4$

Interactive Lab

Move $x$ toward 2. The original rational expression is undefined at the hole, yet its values approach 4. Toggle the simplified form to see the continuous line that fills the gap.

Value: —
x y
Original: (x² − 4)/(x − 2)
At current x the value is shown above. Slide toward 2 to watch the approach.

The amber open circle marks the removable discontinuity. After canceling the common factor the green dashed line is continuous everywhere, including at $x=2$.

Key Idea

$$ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f^{*}(x)}{g^{*}(x)} $$

When direct substitution yields $\frac{0}{0}$, factor and cancel the common zero factor. The starred functions are the simplified versions that agree with the original expression everywhere except at the single canceled point. The limit is then obtained by ordinary substitution into the simplified expression.

$a$The target value that produces the indeterminate form
$f,g$Original numerator and denominator (both zero at $a$)
$f^{*},g^{*}$Simplified numerator and denominator after canceling the common factor
$\frac{0}{0}$Indeterminate form that signals a removable discontinuity

Think Further

  1. Why does canceling a common factor change the function at only one point, yet leave the limit unchanged?
  2. If a sensor returns a reading that algebraically looks like $\frac{0}{0}$, what practical step mirrors the cancellation we performed?
  3. Can every indeterminate form of type $\frac{0}{0}$ be resolved by simple factoring, or are there cases that need other algebraic tricks?
Show suggested answers
  1. The expressions agree wherever the canceled factor is nonzero. They differ at only one point, while the limit depends on nearby values.
  2. Remove the shared source of error or recalibrate the measurement model before evaluating the reading again.
  3. No. Some cases require rationalizing, common denominators, identities, substitutions, or later techniques such as L'Hôpital's rule.