How can simplifying an expression reveal a limit hidden by zero over zero?
Quick answer
If substituting the target value into a fraction gives 0/0, factor the numerator and denominator and look for a shared factor. Cancel it only for inputs near the target that are not the target itself. Then evaluate the simplified expression’s limit. For example, (x²−9)/(x−3) simplifies to x+3 when x≠3, so its limit as x approaches 3 is 6. The original fraction remains undefined at x=3. [S1]
The question, the details, the next step.
I substituted the number into a limit and got zero over zero. Does that mean the limit is zero, or that it doesn’t exist?
Neither conclusion follows just from getting 0/0. It means direct substitution has not told you the limit. For a fraction of polynomials, try factoring the top and bottom and checking for a common factor. If one cancels, the resulting expression may be easier to evaluate near the target. The key is that the original and simplified expressions agree for nearby inputs where the original is defined—not necessarily at the target itself. [S1]
For instance, consider the limit of (x²−9)/(x−3) as x approaches 3. Substituting 3 gives (9−9)/(3−3)=0/0, so we need another step.
How does factoring help in that example? Could you show each algebra step?
Factor the difference of squares in the numerator:
- x²−9=(x−3)(x+3).
- So the fraction is ((x−3)(x+3))/(x−3).
- For x≠3, cancel the shared factor x−3. The expression becomes x+3.
- Now let x approach 3 in the simplified expression: x+3 approaches 6.
Therefore, the limit is 6. This is the same kind of factoring-and-canceling method used for polynomial fractions with a 0/0 substitution result. [S1]
But if we canceled x−3, haven’t we changed the original function at x=3?
The cancellation does not define the original fraction at x=3. At that input, its denominator is zero, so the original expression is undefined. The equality between the original fraction and x+3 holds only when x≠3. A limit asks what values the expression approaches as x gets close to 3, not what the original expression equals at 3. So the nearby values can approach 6 even though the original expression has no value there. [S1]
It can help to write the restriction beside the cancellation: ((x−3)(x+3))/(x−3)=x+3, for x≠3. That qualification prevents the algebra step from being mistaken for filling in a missing function value.
What if there isn’t a factor that cancels? How can I check that I’ve used this method correctly?
First check that your factoring is correct and that you substituted the target accurately. If there is no common factor to cancel, this method has not resolved the limit; don’t force a cancellation or assume the answer. The cited method applies to polynomial numerator and denominator when factoring reveals a common factor. [S1] The example establishes how that method works, but it does not provide a rule for every possible 0/0 expression.
A useful written solution keeps the reasoning visible:
- State the target and show that substitution gives 0/0.
- Factor the numerator and denominator.
- Cancel only a common factor, recording that the cancellation is valid away from the target.
- Evaluate the limit of the simplified expression.
- State the limit separately from whether the original expression is defined at the target.
For this example, the finished result is: “For x≠3, (x²−9)/(x−3)=x+3. Thus, as x approaches 3, the expression approaches 6. The original fraction is undefined at x=3.” [S1]
A practical checklist
Substitute the target value to see whether the expression gives 0/0.
Factor the numerator and denominator to spot shared factors.
Cancel a common factor only while noting that the input is not the target.
Find the limit using the simplified expression.
Report the limit and the original expression’s value or undefined status separately.
A few more useful answers
Does getting 0/0 prove the limit exists?
No. It signals that direct substitution alone did not determine the limit. In the polynomial-fraction method described here, factoring may reveal a cancellable common factor; the 0/0 result itself is not the answer. [S1]
After cancellation, can I say the original function equals the simplified one at the target?
No. In the example, cancellation is valid for x≠3; the original fraction remains undefined at x=3. The limit describes nearby behavior, not a newly assigned value at that point. [S1]
Sources
References used for this guide.
- [S1] 2.3 The Limit Laws — Calculus Volume 1, OpenStax openstax.org · Checked