1.2 · Evaluating limits: substitution, factoring, and rationalization

Start with substitution and limit laws

Polynomials are continuous at every real input, so a polynomial limit can be found by substitution. A rational function can be evaluated by substitution whenever its denominator is nonzero at the target input. For example, $\displaystyle \lim_{x\to2}(3x^2-x+1)=11$.

When $\displaystyle \lim f(x)=L$ and $\displaystyle \lim g(x)=M$ are finite, the limit of a sum is L + M and the limit of a product is LM. For a quotient the result is $\displaystyle \frac{L}{M}$ only if $\displaystyle M\ne0$. Check this condition before substituting into a quotient.

$$\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{L}{M},\quad M\ne0$$

Treat 0/0 as a signal to investigate

A substitution result of $\displaystyle \frac{0}{0}$ is indeterminate: it does not name a limit and does not prove that a limit fails to exist. Different functions producing $\displaystyle \frac{0}{0}$ can have different limits. You need an equivalent expression for nearby inputs.

If a common factor causes the zero denominator, factor numerator and denominator, state the excluded input, and cancel only the common factor. You may cancel factors in a product, never isolated terms across addition.

$$\frac{x^2-9}{x-3}=x+3,\quad x\ne3$$

Use a conjugate for square-root differences

When a difference containing a square root gives $\displaystyle \frac{0}{0}$, multiply numerator and denominator by its conjugate. The identity $\displaystyle (u-v)(u+v)=u^2-v^2$ removes the square-root difference. Then look for a common factor.

For $\displaystyle \frac{\sqrt{x+5}-3}{x-4}$, the conjugate is $\displaystyle \sqrt{x+5}+3$. The numerator becomes x − 4. After cancellation for $\displaystyle x\ne4$, the expression becomes $\displaystyle \frac{1}{\sqrt{x+5}+3}$, whose limit at 4 is $\displaystyle \frac{1}{6}$.

$$\frac{\sqrt{x+5}-3}{x-4}=\frac{1}{\sqrt{x+5}+3},\quad x\ne4$$

A method-selection checklist

First substitute. If all operations are defined, evaluate. If you obtain $\displaystyle \frac{0}{0}$, inspect the expression: polynomial differences suggest factoring; a square-root difference suggests a conjugate; a complex fraction suggests combining fractions. Simplify, then substitute again.

If the numerator approaches a nonzero value while the denominator approaches zero, cancellation is not automatically available. Analyze signs from each side. For example, $\displaystyle \frac{1}{x-2}$ approaches negative infinity from the left and positive infinity from the right, so it has no two-sided limit.

Explain why the new expression is legitimate

A limit concerns nearby inputs with $\displaystyle x\ne a$. Two functions that agree throughout a sufficiently small deleted neighborhood of a have the same limit there. This is why cancellation can compute a limit even though it does not define the original expression at a.

Keep the limit notation until the final evaluation, record the excluded input, and end with a value or a justified statement that the limit does not exist. These steps make your reasoning visible.