Lesson 3: Finding Limits Analytically

(a) Opening Question

Try this one before reading on. Evaluate:

lim_{x→3} (x² − 9)/(x − 3)

Your first instinct might be to plug in x = 3. Do it: the numerator becomes 9 − 9 = 0 and the denominator becomes 3 − 3 = 0. You get 0/0.

Here is the key idea of this entire lesson: 0/0 is not an answer — it is a signal. It tells you the function has a removable problem at x = 3, and that with a little algebra the limit will reveal itself.

Notice that x² − 9 = (x − 3)(x + 3). So for every x ≠ 3:

(x² − 9)/(x − 3) = (x − 3)(x + 3)/(x − 3) = x + 3

The limit only cares about x near 3, not x = 3 itself, so we may cancel that common factor. Then lim_{x→3} (x + 3) = 6. Hold onto that number — and that move. By the end of this lesson you will have three reliable techniques for turning 0/0 into a real answer.

(b) Core Concepts

The Limit Laws

If lim_{x→c} f(x) and lim_{x→c} g(x) both exist, then limits behave exactly the way you would hope. Writing L = lim_{x→c} f(x) and M = lim_{x→c} g(x):

Sum/Difference:  lim (f ± g) = L ± M
Constant multiple: lim (k·f) = k·L
Product:         lim (f·g) = L·M
Quotient:        lim (f/g) = L/M    (provided M ≠ 0)
Power:           lim (f)ⁿ = Lⁿ
Root:            lim ⁿ√f = ⁿ√L      (L ≥ 0 for even n)

These laws are the engine behind everything else. They are also why the techniques below work: each technique rewrites a messy expression into one where the laws do apply.

Direct Substitution — and When It Works

The fastest way to evaluate a limit is direct substitution: plug x = c into the function. The reason this is allowed is continuity.

Direct Substitution Property. If f is continuous at x = c, then lim_{x→c} f(x) = f(c).

Which functions are continuous everywhere on their domain? Essentially all the friendly ones: polynomials, rational functions (everywhere the denominator is nonzero), root functions, exponential and logarithmic functions, and trigonometric functions (on their domains). For these, "take the limit" just means "evaluate."

lim_{x→2} (x³ − 4x + 1) = 8 − 8 + 1 = 1     ✓ polynomial, substitute
lim_{x→0} cos(x) = cos(0) = 1                ✓ trig, substitute
lim_{x→4} (x + 1)/(x − 1) = 5/3              ✓ denominator ≠ 0, substitute

Direct substitution fails only when it produces something undefined. The most important failure is the indeterminate form 0/0 — and that is precisely where the real work begins. (A form like 5/0 is not indeterminate; it signals an infinite limit or a vertical asymptote, the subject of Lesson 2's unbounded behavior.)

Resolving 0/0, Technique 1: Factoring

When substitution gives 0/0 in a rational expression, the numerator and denominator share a common factor of (x − c). Factor, cancel, then substitute. Fully worked:

lim_{x→2} (x² + x − 6)/(x − 2)

Substitute: (4 + 2 − 6)/(2 − 2) = 0/0   → indeterminate
Factor:     (x + 3)(x − 2)/(x − 2)
Cancel:     x + 3        (valid for x ≠ 2)
Substitute: 2 + 3 = 5

So lim_{x→2} (x² + x − 6)/(x − 2) = 5. Notice we canceled (x − 2) before substituting — the whole point is to remove the factor that was causing 0/0.

Resolving 0/0, Technique 2: Rationalizing (Conjugates)

When a square root appears, factoring usually won't help. Instead, multiply by the conjugate over itself. The conjugate of √A − B is √A + B; multiplying them uses the difference of squares (√A − B)(√A + B) = A − B², which clears the root. Fully worked:

lim_{x→0} (√(x + 4) − 2)/x

Substitute: (√4 − 2)/0 = (2 − 2)/0 = 0/0   → indeterminate

Multiply by conjugate (√(x+4) + 2)/(√(x+4) + 2):

  (√(x+4) − 2)(√(x+4) + 2)         (x + 4) − 4              x
  ───────────────────────────  =  ─────────────────  =  ──────────────
        x · (√(x+4) + 2)           x · (√(x+4) + 2)      x(√(x+4) + 2)

Cancel x:   1/(√(x+4) + 2)        (valid for x ≠ 0)
Substitute: 1/(√4 + 2) = 1/(2 + 2) = 1/4

So lim_{x→0} (√(x + 4) − 2)/x = 1/4. The conjugate trick converts a stubborn root into a clean difference of squares, after which the offending x cancels.

Resolving 0/0, Technique 3: Common Denominator

When the expression is a complex fraction (a fraction containing fractions), combine the small fractions over a common denominator first, then simplify.

lim_{x→0} ( 1/(x + 2) − 1/2 ) / x

Substitute: (1/2 − 1/2)/0 = 0/0   → indeterminate

Combine top over common denom 2(x+2):
   1/(x+2) − 1/2 = [2 − (x + 2)] / [2(x + 2)] = (−x) / [2(x + 2)]

Divide by x (multiply by 1/x):
   (−x) / [2(x + 2)] · (1/x) = −1 / [2(x + 2)]      (valid for x ≠ 0)

Substitute: −1/[2(0 + 2)] = −1/4

So lim_{x→0} (1/(x+2) − 1/2)/x = −1/4.

The Squeeze (Sandwich) Theorem

Some limits resist all algebra — especially those involving sin(1/x) or similar bounded-but-wild pieces. For these we trap the function between two others.

Squeeze Theorem. Suppose that for all x in an open interval containing c (except possibly at c itself),

`

g(x) ≤ f(x) ≤ h(x).

`

If lim_{x→c} g(x) = lim_{x→c} h(x) = L, then lim_{x→c} f(x) = L.

In words: if f is sandwiched between two functions that meet at the same height L, then f is forced to that height too. The hypotheses matter: you need the inequality near c and the two outer limits must be equal.

Mini-example. Evaluate lim_{x→0} x² sin(1/x). Since −1 ≤ sin(1/x) ≤ 1, multiplying by x² ≥ 0 gives −x² ≤ x² sin(1/x) ≤ x². Both outer limits are lim_{x→0}(±x²) = 0, so by the Squeeze Theorem the limit is 0.

The Special Trig Limits

The Squeeze Theorem proves the single most important limit in trigonometry:

lim_{x→0} (sin x)/x = 1          (x in radians)

• A smooth wave-like curve that flattens toward the x-axis far from the origin
• A hole (removable discontinuity) at (0, 1): the function is undefined at x = 0 but approaches height 1
• Symmetric about the y-axis (even function)
• The curve hugs the line y = 1 closely near x = 0, confirming the limit equals 1]

Geometrically, for small x the arc length, the chord, and the tangent segment all become nearly equal, which forces (sin x)/x → 1. The companion limit is:

lim_{x→0} (1 − cos x)/x = 0

This one follows from the first: multiply by the conjugate (1 + cos x) to get (1 − cos²x)/[x(1 + cos x)] = (sin²x)/[x(1 + cos x)] = (sin x/x)·(sin x)/(1 + cos x), whose limit is 1 · 0/2 = 0.

Why they matter: these will reappear constantly when we differentiate sin x and cos x in Unit 2. To use them, the argument of sine must match the denominator. For example, lim_{x→0} (sin 5x)/x is not 1 — rewrite it as 5 · (sin 5x)/(5x) = 5 · 1 = 5.

Limits at Infinity of Rational Functions

For lim_{x→±∞} of a rational function p(x)/q(x), compare the degrees (or divide every term by the highest power of x in the denominator). Let n = degree of numerator, m = degree of denominator:

n < m  →  limit is 0          (denominator wins)
n = m  →  limit is the ratio of leading coefficients
n > m  →  limit is ±∞ (no finite limit; numerator wins)

Worked (n = m). lim_{x→∞} (3x² − 2x + 1)/(5x² + 7). Divide every term by x²:

(3 − 2/x + 1/x²)/(5 + 7/x²) → (3 − 0 + 0)/(5 + 0) = 3/5

Each 1/xᵏ term vanishes as x → ∞, leaving the ratio of leading coefficients, 3/5.

(c) Worked Examples

Example 1 — Factor (foundation) · [NO CALC]

Problem. Evaluate lim_{x→2} (x³ − 8)/(x − 2).

Strategy. Substitution gives (8 − 8)/(2 − 2) = 0/0. Factor the numerator as a difference of cubes: a³ − b³ = (a − b)(a² + ab + b²).

Solution.

x³ − 8 = (x − 2)(x² + 2x + 4)
(x³ − 8)/(x − 2) = x² + 2x + 4        (x ≠ 2)
lim_{x→2} (x² + 2x + 4) = 4 + 4 + 4 = 12

Answer: 12. Verification: the factor (x² + 2x + 4) is continuous, so direct substitution after canceling is valid.

Example 2 — Rationalize (procedural) · [NO CALC]

Problem. Evaluate lim_{x→3} (√(x + 1) − 2)/(x − 3).

Strategy. Substitution gives (√4 − 2)/0 = 0/0. A root sits on top, so multiply by the conjugate √(x + 1) + 2.

Solution.

(√(x+1) − 2)(√(x+1) + 2)        (x + 1) − 4            x − 3
─────────────────────────  =  ──────────────────  =  ──────────────────
   (x − 3)(√(x+1) + 2)         (x − 3)(√(x+1) + 2)    (x − 3)(√(x+1) + 2)

Cancel (x − 3):  1/(√(x+1) + 2)       (x ≠ 3)
Substitute x = 3:  1/(√4 + 2) = 1/(2 + 2) = 1/4

Answer: 1/4. Justification: canceling (x − 3) is valid because the limit ignores x = 3 itself, and the result is continuous at x = 3.

Example 3 — Trig limit using (sin x)/x (AP level) · [NO CALC]

Problem. Evaluate lim_{x→0} (sin 5x)/(sin 3x).

Strategy. Both numerator and denominator → 0, giving 0/0. We want to manufacture two copies of the pattern (sin ▢)/▢, one for 5x and one for 3x.

Solution. Multiply and divide to match arguments:

sin 5x     sin 5x     5x      3x
──────  =  ──────  ·  ──  ·  ──────
sin 3x       5x       3x     sin 3x

       =  (sin 5x / 5x) · (5x/3x) · (3x / sin 3x)

As x → 0: (sin 5x)/(5x) → 1, (3x)/(sin 3x) → 1, and 5x/3x = 5/3 exactly.

lim_{x→0} (sin 5x)/(sin 3x) = 1 · (5/3) · 1 = 5/3

Answer: 5/3. Key move: each sine's argument must match its denominator, so we insert 5x under the first sine and 3x under the second.

Example 4 — Limit at infinity (AP level) · [NO CALC]

Problem. Evaluate lim_{x→∞} √(4x² + 1)/(x + 1).

Strategy. Both top and bottom grow without bound. Inside the root the highest power is x², so the numerator behaves like √(4x²) = 2|x| = 2x for large positive x. Divide top and bottom by x — and remember that for x > 0, x = √(x²).

Solution.

√(4x² + 1)        √(4x² + 1)/√(x²)        √(4 + 1/x²)
───────────  =  ────────────────────  =  ──────────────       (x > 0)
   x + 1            (x + 1)/x               1 + 1/x

As x → ∞:  √(4 + 0) / (1 + 0) = √4 / 1 = 2

Answer: 2. Caution: this works because x → +∞ makes x > 0, so √(x²) = x. For x → −∞, √(x²) = −x, and the answer would flip sign to −2.

(d) Common Mistakes

Substituting into 0/0 and stopping.

What students do: write "= 0/0" and call it the answer (or call it 0, or undefined).

Why it's wrong: 0/0 is indeterminate — it has no fixed value and is a prompt to do algebra.

Fix: treat 0/0 as a green light for factoring, rationalizing, or common denominators.

Multiplying by the conjugate but forgetting to multiply the denominator.

What students do: multiply only the numerator by the conjugate.

Why it's wrong: that changes the value of the expression. You must multiply by (conjugate)/(conjugate) = 1.

Fix: multiply top and bottom by the same conjugate; expand only the part with the root and leave the other factor unmultiplied so an (x − c) can cancel.

Forgetting that (sin ▢)/▢ requires matching arguments.

What students do: claim lim_{x→0} (sin 5x)/x = 1.

Why it's wrong: the rule is (sin u)/u → 1 only when the same u is on top and bottom. Here the bottom is x, not 5x.

Fix: rewrite (sin 5x)/x = 5·(sin 5x)/(5x) → 5·1 = 5.

Degree-comparison errors at infinity.

What students do: read off the constant terms, or guess "the limit is the leading coefficients" even when degrees differ.

Why it's wrong: only the highest-degree terms control behavior at infinity, and only when n = m do you take the ratio of leading coefficients.

Fix: compare degrees first — n, n=m → ratio of leads, n>m → ±∞ — or divide every term by the highest power in the denominator.

(x − 4)(√x + 2)       (x − 4)(√x + 2)
────────────────  =  ─────────────────  = √x + 2     (x ≠ 4)
(√x − 2)(√x + 2)          x − 4
lim_{x→4} (√x + 2) = √4 + 2 = 4

(sin 5x)/(sin 3x) = [(sin 5x)/(5x)]·[(3x)/(sin 3x)]·(5x/3x)
                  → 1 · 1 · (5/3) = 5/3

−x² ≤ x² cos(1/x) ≤ x².

(x³ + 1)/(x² + 1) = (x + 1/x²)/(1 + 1/x²) → (x + 0)/(1 + 0) = x,