Recipe

Evaluate a limit algebraically

Use when: you're asked for \lim_{x\to a}f(x) and f is built from polynomials, radicals, or rational expressions

Goal: a single number (or a clear reason the limit doesn't exist), reached without a table or a graph

Taught in Algebraic techniques for limits

1. 1

   Plug x=a straight into f(x).

   • the result is a real number — that number IS the limit; stop here.
   • the result is 0/0 — continue to Step 2, the "removable hole" signal.
   • the result is a nonzero number over 0 — the limit is infinite (or doesn't exist); that's a different technique, covered a few sections ahead.

   Why: a limit only fails to equal f(a) when f has a genuine gap or jump at a, and direct substitution is the fastest way to rule that out.

2. 2

   If direct substitution gave 0/0 and f is a rational expression, factor the numerator and denominator, then cancel whatever factor is forcing both to vanish at x=a.

   Example

   For \lim_{x\to3}\dfrac{x^2-9}{x-3}: substituting x=3 gives 0/0. Factoring, \dfrac{x^2-9}{x-3}=\dfrac{(x-3)(x+3)}{x-3}=x+3 for x\neq3.

   Why: the factor causing the 0/0 is exactly what canceling removes — it was never really part of the function's behavior near a, only at a itself.

3. 3

   If the 0/0 expression has a square root, multiply top and bottom by the conjugate of the radical piece instead of trying to factor it.

   Example

   For \lim_{x\to4}\dfrac{\sqrt{x}-2}{x-4}: multiply by \dfrac{\sqrt x+2}{\sqrt x+2} to get \dfrac{x-4}{(x-4)(\sqrt x+2)}=\dfrac{1}{\sqrt x+2}.

   Why: a conjugate turns a difference of square roots into a difference of squares, exactly the algebra a factoring approach can't reach.

4. 4

   Plug x=a into whatever's left after Step 2 or Step 3.

   Why: the simplified expression agrees with the original everywhere except at x=a itself, and a limit never cares about that one point.