Algebraic Limits & the Squeeze Theorem

You’ll be able to

Evaluate limits by direct substitution and recognize the 0/0 indeterminate form
Resolve 0/0 limits by factoring, rationalizing, or simplifying
Apply the Squeeze Theorem to trap a limit between two known bounds

Direct substitution first

For any function built from polynomials, roots, and the standard trig, exponential, and log functions, the fastest move is direct substitution: plug x = a in. If you get a finite number, that number is the limit, because these functions are continuous wherever they are defined. Substitution is not a shortcut to be ashamed of — it is the correct method whenever it produces a real value.

The 0/0 indeterminate form

Substitution sometimes gives 0/0, which is called an indeterminate form. It does not mean the limit is 0, or 1, or nonexistent — it means substitution cannot decide, and you must do algebra first. The usual rescues are to factor and cancel the common term causing the zero, or to rationalize (multiply by a conjugate) when a square root is involved. After the offending factor cancels, substitute again.

The Squeeze Theorem

Some limits resist algebra — especially oscillating ones like sin(1/x). The Squeeze Theorem (also called the Sandwich Theorem) handles them: if g(x) ≤ f(x) ≤ h(x) near a, and the two outer functions share a limit, lim(x→a) g(x) = lim(x→a) h(x) = L, then f is trapped and forced to the same value, lim(x→a) f(x) = L. The idea is to bound a messy function between two tame ones that meet.

Squeeze Theorem

If g(x) ≤ f(x) ≤ h(x) near a and lim(x→a) g(x) = lim(x→a) h(x) = L, then lim(x→a) f(x) = L

The classic use: since −x² ≤ x²·sin(1/x) ≤ x² and both bounds → 0, the middle → 0 as well.

A special trig limit

lim(x→0) sin(x)/x = 1

Proved with the Squeeze Theorem. Memorize it; it underlies the derivative of sin(x). Note x must be in radians.

Worked example

Evaluate lim(x→4) (x² − 16)/(√x − 2).

1.Substitute x = 4: numerator 16 − 16 = 0, denominator 2 − 2 = 0 — the 0/0 indeterminate form.
2.Rationalize by multiplying top and bottom by the conjugate (√x + 2).
3.The denominator becomes (√x − 2)(√x + 2) = x − 4, and the numerator becomes (x² − 16)(√x + 2) = (x − 4)(x + 4)(√x + 2).
4.Cancel the common (x − 4): the expression simplifies to (x + 4)(√x + 2).
5.Substitute x = 4: (4 + 4)(√4 + 2) = 8 × 4 = 32.

Answer: lim(x→4) (x² − 16)/(√x − 2) = 32.

Watch out

A 0/0 result is a signal to do more work, not an answer. Writing "the limit is 0/0, so it does not exist" is wrong — most 0/0 limits have a finite value that appears only after you factor or rationalize.

Checkpoint

Evaluate lim(x→2) (x² + x − 6)/(x − 2).

Checkpoint

Given that 3 − x² ≤ f(x) ≤ 3 + x² for all x near 0, what is lim(x→0) f(x)?

On the exam

Reach for the Squeeze Theorem whenever you see a bounded factor (like sin or cos of something) multiplied by a factor going to 0. Bound the trig part between −1 and 1, multiply through, and let the outer limits pinch the answer.

Answer the 2 checkpoints as you read.