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Algebraic Limit Techniques · Tutorial 31 of 999

Why Cancelling a Factor Preserves a Limit

See why cancellation preserves nearby values, how it reveals limits at removable holes, and what conditions must still be checked.

Beginner 9 min read

What You'll Learn

  • Explain why a function’s value at the target does not determine its limit there
  • State when two expressions that agree near a target have the same limit
  • Cancel a shared factor to evaluate limits at removable holes
  • Check whether the simplified denominator is nonzero at the target
  • Identify why cancellation cannot be performed at the point where the shared factor is zero
  • Recognize when cancellation alone does not resolve a limit

Why a Hole Does Not Determine a Limit

A limit describes the values of a function near a target, not necessarily the value of the function at the target itself. This distinction explains why cancelling a shared factor can reveal a limit even when the original expression is undefined at the point we are approaching.

For example, the expression \(\frac{x^2-4}{x-2}\) is undefined at \(x=2\), because its denominator is zero there. But for nearby values \(x\ne 2\), its numerator and denominator have a shared factor. Cancelling that factor gives an expression whose values are easy to examine. The cancellation does not make the original expression defined at \(x=2\); it shows that the two expressions agree at all sufficiently nearby points other than \(2\).

Key idea: A limit is unchanged if an expression is replaced by another expression that has the same values for all sufficiently nearby inputs except possibly at the target.

What It Means to Agree Near a Target

Suppose \(f\) and \(g\) are functions defined for inputs near \(a\), perhaps with either function undefined at \(a\). They agree on a punctured neighborhood of \(a\) if there is some positive distance around \(a\) such that \(f(x)=g(x)\) whenever \(x\) is within that distance but \(x\ne a\).

The phrase “but \(x\ne a\)” matters. A limit considers values approaching \(a\), not the value at \(a\). Thus, two functions can have the same limit even if one or both are undefined at \(a\), or if their values at \(a\) differ.

Theorem: If \(f(x)=g(x)\) for every \(x\) in some punctured neighborhood of \(a\), and \(\lim_{x\to a}g(x)=L\), then \(\lim_{x\to a}f(x)=L\) as well. The value of either function at \(a\), if it is defined, does not affect this conclusion.

This theorem is the reason cancellation preserves a limit. Cancelling a factor produces an expression equal to the original one at every nearby input where the cancellation is valid. The limit can then be found using the simpler expression.

Why the Theorem Works

Here is a justification using the definition of a limit. The statement \(\lim_{x\to a}g(x)=L\) means that for every positive tolerance \(\varepsilon\), there is a positive distance \(\delta\) such that whenever \(0<|x-a|<\delta\), the value \(g(x)\) is within \(\varepsilon\) of \(L\).

Choose \(\delta\) small enough that \(f(x)=g(x)\) whenever \(0<|x-a|<\delta\). Since \(g\) tends to \(L\), for any given \(\varepsilon>0\) we can also choose a distance that ensures \(|g(x)-L|<\varepsilon\). Use the smaller of these distances. Then \(f(x)=g(x)\), so \(f(x)\) is also within \(\varepsilon\) of \(L\). This proves that \(f\) tends to \(L\).

$$ 0<|x-a|<\delta \quad\Longrightarrow\quad f(x)=g(x) \quad\Longrightarrow\quad |f(x)-L|=|g(x)-L|<\varepsilon. $$

In an algebraic limit problem, the equality \(f(x)=g(x)\) usually comes from cancelling a factor. The restriction \(x\ne a\) is essential: if the shared factor is zero at \(a\), division by it is not valid at \(a\). Limits use the nearby equality, not an invalid substitution at the target.

Worked Examples: Cancel and Evaluate

Worked Example: A Difference of Squares

Evaluate \(\lim_{x\to 2}\frac{x^2-4}{x-2}\).

Direct substitution into the original expression gives numerator \(2^2-4=0\) and denominator \(2-2=0\). The expression is undefined at \(x=2\), so this substitution does not evaluate the limit. Factor the numerator:

$$ x^2-4=(x-2)(x+2). $$

For \(x\ne 2\), cancel the shared factor \(x-2\):

$$ \frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x+2. $$

The original expression and \(x+2\) agree at every nearby \(x\ne 2\). Therefore, the limit equals the limit of \(x+2\). Substituting \(2\) into this simplified expression gives \(2+2=4\).

Check: Multiplying the factors back gives \((x-2)(x+2)=x^2-4\), the original numerator. The simplified expression approaches \(4\) as \(x\to 2\), while the original quotient remains undefined at \(x=2\). Thus the limit is \(4\), not a value of the original function at \(2\).

Worked Example: A Cubic Difference

Evaluate \(\lim_{x\to 2}\frac{x^3-8}{x-2}\).

For \(x\ne 2\), the difference-of-cubes factorization gives \(x^3-8=(x-2)(x^2+2x+4)\). To check the factorization, multiply:

$$ (x-2)(x^2+2x+4) =x^3+2x^2+4x-2x^2-4x-8 =x^3-8. $$

Now cancel \(x-2\), which is nonzero for the nearby inputs under consideration:

$$ \frac{x^3-8}{x-2} = \frac{(x-2)(x^2+2x+4)}{x-2} = x^2+2x+4 \qquad (x\ne 2). $$

The simplified polynomial has limit \(2^2+2(2)+4=4+4+4=12\) as \(x\to 2\). Therefore, the original quotient has limit \(12\).

Check: At a nearby value such as \(x=1\), the original quotient is \(\frac{1-8}{1-2}=\frac{-7}{-1}=7\), and the simplified expression is \(1+2+4=7\). The expressions agree where cancellation is valid, and the simplified expression tends to \(12\) at the target.

Worked Example: Cancelling When a Denominator Remains

Evaluate \(\lim_{x\to 3}\frac{(x-3)(x+1)}{(x-3)(x+2)}\).

The original numerator and denominator are both zero at \(x=3\), so the original quotient is undefined there. For \(x\ne 3\), cancel the shared factor:

$$ \frac{(x-3)(x+1)}{(x-3)(x+2)} = \frac{x+1}{x+2}. $$

The remaining denominator is nonzero at the target: \(3+2=5\). Thus the quotient limit law, or direct evaluation of this continuous expression where its denominator is nonzero, gives

$$ \lim_{x\to 3}\frac{x+1}{x+2} = \frac{3+1}{3+2} = \frac{4}{5}. $$

Check: At \(x=4\), the original quotient is \(\frac{(1)(5)}{(1)(6)}=\frac{5}{6}\), and the simplified quotient is \(\frac{5}{6}\). They agree at this nearby input. Since the simplified denominator approaches \(5\), not \(0\), its limit is indeed \(\frac45\).

Conditions and Limits of Cancellation

Cancellation is an equality of expressions only where the cancelled factor is nonzero. If the factor vanishes at the target, that usually still leaves a whole punctured neighborhood where cancellation is valid. For example, \(x-a\ne 0\) whenever \(x\ne a\). The limit theorem then transfers the limit from the simplified expression to the original one.

After cancellation, check what remains. If the simplified expression is a polynomial, its limit can be found by evaluation. If it is a quotient, its denominator must be nonzero at the target before direct evaluation using the quotient limit law. Cancellation by itself does not guarantee a finite limit or even settle the limit.

Worked Example: Cancellation Does Not Always Finish the Problem

Consider \(\lim_{x\to 0}\frac{x^2}{x^2}\). For \(x\ne 0\), cancellation gives

$$ \frac{x^2}{x^2}=1. $$

The simplified expression is constant, so its limit is \(1\). The original expression is undefined at \(0\), but it equals \(1\) for every nearby nonzero \(x\). Thus the limit is \(1\).

Now compare \(\lim_{x\to 0}\frac{x^2}{x^3}\). Cancelling \(x^2\) for \(x\ne 0\) gives \(\frac{1}{x}\). The denominator still approaches \(0\), and the simplified expression does not approach a finite number: its values are positive and grow without bound from the right, and negative and decrease without bound from the left. Cancellation exposed the remaining behavior but did not produce a finite limit.

Check: For \(x\ne0\), \(x^2/x^3=1/x\) exactly. At \(x=0.1\), both expressions equal \(10\); at \(x=-0.1\), both equal \(-10\). The opposite one-sided behavior confirms that the two-sided limit does not exist.

A Common Mistake: Cancelling at the Target

A tempting but incorrect argument is to substitute the target first and treat the shared zero factor as if it can be cancelled there. For instance, someone might write

$$ \frac{2^2-4}{2-2} = \frac{(2-2)(2+2)}{2-2} = 2+2 = 4. $$

This is not a valid evaluation: the first quotient has denominator \(0\), and the cancellation step divides by \(2-2=0\). The correction is to state the restriction before cancelling:

$$ \frac{x^2-4}{x-2}=x+2 \qquad\text{only for }x\ne2. $$

Then use the equality for nearby inputs other than \(2\) to evaluate the limit. The answer \(4\) is the limit, not the value of the original quotient at \(2\).

Another related error is cancelling part of a sum rather than a factor. Cancellation applies to common factors of a numerator and denominator, not to terms that merely appear in both. Always factor the expressions and check the equality after cancelling.

Why Removable Holes Matter

A removable hole occurs when an expression is undefined at a point but has a finite limit there. The cancelled expression describes the nearby values and supplies the limit; it does not automatically fill in the missing function value. For example, \(\frac{x^2-4}{x-2}\) has a limit of \(4\) at \(x=2\), although the original quotient has no value there. A new function could be defined to equal \(4\) at \(2\), but that would be a separate definition.

This distinction is useful throughout limit work: first establish where two formulas agree, then use the simpler formula to study nearby behavior, and finally keep the limit separate from any value assigned at the target.

Check Your Understanding

For each question, pay attention to where cancellation is valid and whether the simplified expression has a limit.

  1. For \(\frac{x^2-9}{x-3}\), what expression agrees with the quotient when \(x\ne3\), and what is the limit as \(x\to3\)?
  2. Why does cancelling a factor that is zero at the target not establish an equality at the target?
  3. If two functions agree whenever \(0<|x-a|<\delta\), why must they have the same limit at \(a\), provided one limit exists?
  4. After cancelling a shared factor from a quotient, what must be checked before using direct substitution in the simplified quotient?
  5. What does cancelling in \(\frac{x^2}{x^3}\) reveal about the limit as \(x\to0\)?

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