Lesson 5: Limits Unit Review

(a) Opening Question

Before you read anything, try this multi-skill warm-up. It quietly touches five different Unit 1 ideas. No calculator.

Let

        ⎧  (x² - 4)/(x - 2),     x < 2
f(x) =  ⎨  5,                    x = 2
        ⎩  (3x² + 1)/(x² - 1),   x > 2

(i) Find lim_{x→2⁻} f(x).

(ii) Find lim_{x→2⁺} f(x).

(iii) Does lim_{x→2} f(x) exist? If so, what is it?

(iv) Is f continuous at x = 2? Name the type of any discontinuity.

(v) Find lim_{x→∞} f(x).

Take two minutes. Jot answers before peeking. (i) is a factor-and-cancel 0/0. (ii) is a direct substitution that does not hit a zero denominator. (iii) compares the two one-sided results. (iv) checks the three-part continuity definition against f(2) = 5. (v) is a limit at infinity — compare leading-degree behavior. Full solution lives in the Answer Key as Warm-Up.

(b) Core Concepts — The Big Picture of Unit 1

You have learned a pile of separate techniques. The exam does not tell you which one to use — recognizing the situation is the skill. This section is a synthesis, not a re-teach. Keep it as your one-page map.

The limit decision guide

To evaluate lim_{x→c} f(x), walk this ladder top to bottom and stop at the first rule that works.

Try direct substitution. If f is built from polynomials, rationals, roots, trig, exponential, or log functions and c is in the domain, then lim_{x→c} f(x) = f(c). This is the definition of continuity working for you. Most limits that "look hard" actually substitute cleanly.

Read the form you get.

- A finite number → that's the answer. Done.

- (nonzero)/0 → the limit is infinite (±∞) or does not exist. Check the sign of the denominator from each side to decide +∞, −∞, or DNE. This signals a vertical asymptote.

- 0/0 → indeterminate. Substitution failed; the limit may still exist. Go to step 3.

Resolve a 0/0 form with one of three algebraic moves:

- Factor and cancel the common factor causing the zero. (e.g. (x²−4)/(x−2) = x+2.)

- Rationalize when a square root is present — multiply by the conjugate. (e.g. (√(x+1)−1)/x.)

- Combine/simplify complex fractions or use a known special limit:

lim_{x→0} (sin x)/x = 1        lim_{x→0} (1 − cos x)/x = 0

(Both require x in radians. The second is often written lim_{x→0} (cos x − 1)/x = 0.)

Limits at infinity (x→±∞) for a rational function: compare the degrees of numerator (n) and denominator (d).

- n < d → limit is 0 (horizontal asymptote y = 0).

- n = d → limit is the ratio of leading coefficients (horizontal asymptote at that value).

- n > d → limit is ±∞ (no horizontal asymptote; may be a slant asymptote, not tested for value on AB).

A clean trick: divide every term by the highest power of x in the denominator. For roots, remember √(x²) = |x|, so as x→−∞ a sign flip appears.

Squeeze (Sandwich) Theorem when f is trapped between two functions sharing a limit: if g(x) ≤ f(x) ≤ h(x) near c and lim g = lim h = L, then lim f = L. The classic is lim_{x→0} x² sin(1/x) = 0.

### Existence of a limit

lim_{x→c} f(x) = L exists iff the one-sided limits agree:

lim_{x→c⁻} f(x) = lim_{x→c⁺} f(x) = L

A limit can exist even when f(c) is undefined or has a different value. A limit fails to exist when the one-sided limits disagree (jump), when the function grows without bound (infinite), or when it oscillates (e.g. sin(1/x) near 0).

Continuity — the three conditions

f is continuous at x = c iff all three hold:

1.  f(c) is defined
2.  lim_{x→c} f(x) exists
3.  lim_{x→c} f(x) = f(c)

If any condition fails, classify the discontinuity:

• Removable (hole): lim_{x→c} f(x) exists but ≠ f(c) (or f(c) undefined). You could "patch" it by redefining one point.
• Jump: the one-sided limits both exist but are unequal. (Common in piecewise functions.)
• Infinite: one or both one-sided limits are ±∞. Lives at a vertical asymptote.

A function is continuous on an interval if it is continuous at every point inside it (using one-sided continuity at closed endpoints).

The Intermediate Value Theorem (state the conditions!)

IVT. If f is continuous on the closed interval [a, b] and k is any value between f(a) and f(b), then there exists at least one c in (a, b) with f(c) = k.

IVT is an existence theorem — it guarantees a solution exists; it never tells you the value of c or how many. To invoke it on the exam you must (1) state continuity on [a, b], and (2) show k lies strictly between f(a) and f(b). The most common use is proving an equation f(x) = k (often = 0) has a root in an interval.

Must-know facts to have memorized

• lim_{x→0} (sin x)/x = 1 and lim_{x→0} (1 − cos x)/x = 0.
• Polynomials are continuous everywhere; rationals are continuous except where the denominator is 0.
• A removable discontinuity is exactly where numerator and denominator share a factor that cancels.
• Horizontal asymptotes come from limits at ±∞; vertical asymptotes come from (nonzero)/0.
• "Limit exists" ≠ "function is continuous." Continuity is the stronger statement.

(c) Worked Examples

Example 1 — Reading a limit off a graph (NO CALC)

[GRAPH: piecewise f on [-2, 5] × [-1, 6]
- Open circle at (1, 4), closed circle at (1, 2): the curve approaches y=4 from both sides but f(1)=2
- Smooth curve approaching (3, 1) from the left (open circle), and approaching (3, 5) from the right (open circle); no point plotted at x=3
- Vertical asymptote at x = 4: branch to the left rises to +∞, branch to the right drops to −∞
- Otherwise continuous]

Problem. Find: (a) lim_{x→1} f(x) and f(1); (b) lim_{x→3} f(x); (c) lim_{x→4⁻} f(x); (d) classify each discontinuity.

Strategy. Limits care about approach, not the plotted point. Compare left and right at each break.

Solution.

(a) Both sides approach 4, so lim_{x→1} f(x) = 4. But the solid dot says f(1) = 2. Limit exists; value of the function differs.

(b) Left side → 1, right side → 5. They disagree, so lim_{x→3} f(x) does not exist.

(c) The left branch rises without bound: lim_{x→4⁻} f(x) = +∞ (so the two-sided limit DNE and it's an infinite discontinuity).

(d) At x=1: removable (limit exists, ≠ f(1)). At x=3: jump. At x=4: infinite.

Justification (AP style). "At x = 1, lim_{x→1} f(x) = 4 exists because the left- and right-hand limits are both 4, but f(1) = 2 ≠ 4, so f has a removable discontinuity at x = 1."

Example 2 — An analytical 0/0 (NO CALC)

Problem. Evaluate lim_{x→3} (x² − x − 6)/(x² − 9).

Strategy. Substitution gives 0/0 — indeterminate. Factor both.

Solution. Numerator x² − x − 6 = (x − 3)(x + 2). Denominator x² − 9 = (x − 3)(x + 3). Cancel (x − 3):

lim_{x→3} (x + 2)/(x + 3) = (3 + 2)/(3 + 3) = 5/6

Verify. The cancellation is valid for x ≠ 3, which is all that matters for a limit. Substituting in the simplified form is legal because x + 3 → 6 ≠ 0. Answer: 5/6. (Geometrically, the original function has a removable hole at x = 3 and a vertical asymptote at x = −3.)

Example 3 — A limit at infinity (NO CALC)

Problem. Evaluate lim_{x→∞} (√(4x² + 1))/(3x − 5).

Strategy. Top behaves like √(4x²) = 2|x|. For x→+∞, |x| = x. Divide through by x.

Solution. For x > 0, √(4x² + 1) = √(x²)·√(4 + 1/x²) = x·√(4 + 1/x²). Then

(√(4x² + 1))/(3x − 5) = (x√(4 + 1/x²))/(x(3 − 5/x))
                      = √(4 + 1/x²)/(3 − 5/x)

As x→∞, 1/x² → 0 and 5/x → 0, giving √4/3 = 2/3. Answer: 2/3 (horizontal asymptote y = 2/3 to the right).

Watch the sign. As x→−∞, √(x²) = |x| = −x, so that same limit would be −2/3. Different asymptote on the left.

Example 4 — IVT / continuity justification (NO CALC)

Problem. Let f(x) = x³ + x − 1. Show that f has a real root in the interval (0, 1).

Strategy. f is a polynomial → continuous everywhere. Evaluate the endpoints and apply the IVT with k = 0.

Solution & justification. f is a polynomial, so it is continuous on [0, 1]. Compute f(0) = 0 + 0 − 1 = −1 and f(1) = 1 + 1 − 1 = 1. Since f(0) = −1 < 0 < 1 = f(1), the value 0 lies between f(0) and f(1). By the Intermediate Value Theorem, there exists a c in (0, 1) such that f(c) = 0. Therefore f has a real root in (0, 1). ∎

Note the discipline: state continuity on the closed interval, show 0 is strictly between the two endpoint values, then conclude existence on the open interval. Dropping either condition loses points.

(d) Common Mistakes — Unit 1's Top Traps

"The limit equals the function value." Not always. Students plug in x = c reflexively. That only works when f is continuous there. At holes, jumps, and asymptotes the limit and f(c) can differ — or the limit may not exist at all. Fix: always ask "is this point a break?" before substituting.

Treating 0/0 as if it means the limit is 0 (or DNE). 0/0 is indeterminate — it tells you nothing yet. The limit could be any number, ∞, or DNE. Fix: factor, rationalize, or simplify, then re-evaluate.

Forgetting √(x²) = |x| for limits at −∞. Pulling x out of a square root without the absolute value drops a needed sign flip, turning a −2/3 into 2/3. Fix: for x→−∞, replace √(x²) with −x.

Claiming a limit exists when only one side does. A vertical asymptote where the left side goes +∞ and the right side goes −∞ means the two-sided limit does not exist — you cannot report it as +∞. Fix: check both one-sided limits before declaring a two-sided value.

Botching the IVT conditions. Writing "by IVT there's a root" without stating continuity on [a, b], or applying IVT to a function with a discontinuity in the interval. Fix: every IVT argument names continuity on the closed interval and shows the target value is between the endpoints.

---

## (e) Practice Problems

Mix of [CALC] and [NO CALC], spanning all of Unit 1. Items 13–15 are FRQ-style free-response with required justification. Full solutions in the Answer Key.

FRQ-style free-response (justify your work)

Piecewise continuity (4 points). Let

        ⎧  x² + 1,        x < 1
f(x) =  ⎨  a x + b,       1 ≤ x ≤ 3
        ⎩  10 − x,        x > 3

(a) Find lim_{x→1⁻} f(x) and lim_{x→3⁺} f(x). (1 pt)

(b) Find the values of a and b that make f continuous on all of ℝ. Show the equations you solve. (2 pts)

(c) With those values, is f continuous at x = 1? Justify using the three-part definition. (1 pt)

Limits and asymptotes (4 points). Let f(x) = (2x² − 8)/(x² − x − 6).

(a) Factor numerator and denominator and identify all x-values where f is undefined. (1 pt)

(b) Classify the discontinuity at each such x (removable or infinite). Justify with a limit. (2 pts)

(c) Find lim_{x→∞} f(x) and state the horizontal asymptote. (1 pt)

IVT application (3 points). A continuous function g has these measured values:

| x | 0 | 2 | 4 | 6 |

|---|---|---|---|---|

| g(x) | −3 | 1 | −2 | 5 |

(a) Explain why the IVT guarantees at least two solutions to g(x) = 0 on [0, 6], and state intervals containing them. (2 pts)

(b) Does the IVT guarantee that g(x) = 4 has a solution on [0, 4]? Explain. (1 pt)

---

f(x) = (2(x − 2)(x + 2))/((x − 3)(x + 2))