1.6 Determining Limits Using Algebraic Manipulation

Direct substitution and the limit laws settle most limits, but some expressions break down at exactly the input being approached: substitution returns $\frac{0}{0}$, and no limit law applies to the expression as written. Algebraic manipulation rewrites the expression into an equivalent form, valid near the target, whose limit can be found directly.

When algebra must come first

For a function given analytically, first try direct substitution. If it produces a real number, the limit laws often finish the problem. If it produces a form such as $\frac{0}{0}$, the expression has not been evaluated; instead, it signals that algebraic manipulation is needed.

Consider the average-cost expression

$$ C(x)=\frac{x^2-9}{x-3}, $$

where $x$ represents production level. At $x=3$, direct substitution gives

$$ \frac{3^2-9}{3-3}=\frac{0}{0}. $$

Factor the numerator:

$$ x^2-9=(x-3)(x+3). $$

For $x\ne3$,

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

Therefore,

$$ \lim_{x\to3}C(x) =\lim_{x\to3}(x+3)=6. $$

The original function may be undefined at $x=3$, but a limit examines values near $3$, not necessarily the value at $3$.

Rationalizing a difference

A square-root expression can also hide a removable factor. For

$$ \lim_{x\to4}\frac{\sqrt{x}-2}{x-4}, $$

multiplying by the conjugate gives

$$ \frac{\sqrt{x}-2}{x-4}\cdot \frac{\sqrt{x}+2}{\sqrt{x}+2}

\frac{x-4}{(x-4)(\sqrt{x}+2)}

\frac{1}{\sqrt{x}+2}. $$

Thus,

$$ \lim_{x\to4}\frac{\sqrt{x}-2}{x-4}

\frac{1}{2+2} =\frac14. $$

This is algebraic manipulation in action: rewrite the expression into an equivalent form on a punctured neighborhood of the target.

Misconception check

Misconception: “If substitution gives $\frac00$, the limit is zero.” The symbol $\frac00$ is not an answer. It means the numerator and denominator both approach zero, so further reasoning—factoring, rationalizing, or another appropriate algebraic procedure—is required.

AP skill alignment: Topic 1.6 develops Skill 1.C: Identify an appropriate mathematical rule or procedure based on the classification of a given expression. Here the classification comes from what direct substitution produces: a real number means the limit laws finish the problem, while $\frac{0}{0}$ calls for factoring, rationalizing, or another rewriting step first.

Retrieval check

Evaluate without technology:

$$ \lim_{x\to1}\frac{x^2-1}{x-1}. $$

Factor first, cancel only for $x\ne1$, and then substitute. The result is

$$ \boxed{2}. $$