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What You'll Learn

• Read limit notation, evaluate one-sided limits, and state exactly when a limit is nonexistent — a standard answer choice on the CLEP exam.
• Estimate limits from graphs and tables, and compute them exactly with the limit laws and direct substitution.
• Resolve the indeterminate form 0/0 three ways: factoring, multiplying by a conjugate, and combining fractions.
• Use the special trig limits sin(x)/x → 1 and (1 − cos x)/x → 0, and recognize when the Squeeze Theorem applies.
• Get oriented to the CLEP item format: five choices (A)–(E), short computational stems, and a two-section exam structure.

Content

How the CLEP exam asks limit questions

Before the math, know the format. CLEP Calculus runs about 44 questions in roughly 90 minutes, split into two sections: Section 1 (about 27 questions, ~50 minutes) with no calculator, and Section 2 (about 17 questions, ~40 minutes) with an online TI-84 Plus CE graphing calculator built into the exam software. (Some testing centers permit a graphing calculator throughout — confirm your center's policy.) Most questions have five answer choices, (A) through (E), and a few are numeric entry: you type a number into a box instead of choosing a letter.

Limit stems are short and direct — often just an expression ending in "is" or "=":

lim_{x→3} (x² − 9)/(x − 3) is (A) 0 (B) 6 (C) 3 (D) 9 (E) nonexistent

Notice "nonexistent" sitting there as a live option. It is correct often enough that you must genuinely test existence every time, not treat it as a throwaway choice. Limits are about 10% of the exam, but limit machinery powers the other 90%, so this lesson pays rent all course long.

What a limit means

The limit of f(x) as x approaches c is the single value L that f(x) gets arbitrarily close to as x gets arbitrarily close to c (with x ≠ c):

lim_{x→c} f(x) = L

The central idea: a limit describes behavior NEAR c, not the value AT c. The function may be undefined at c (a hole), or defined with a value different from the limit — the limit ignores the point itself and watches the trend.

One-sided limits and the existence rule

• Left-hand limit lim_{x→c⁻} f(x): the value f(x) approaches as x comes in from values less than c.
• Right-hand limit lim_{x→c⁺} f(x): the value approached from values greater than c.

lim_{x→c} f(x) = L   if and only if   lim_{x→c⁻} f(x) = L   AND   lim_{x→c⁺} f(x) = L

A two-sided limit is nonexistent in three classic ways:

1. The one-sided limits disagree (a jump).
2. Unbounded behavior — f(x) → +∞ or −∞ (no finite limit; the graph has a vertical asymptote).
3. Oscillation — the standard example is lim_{x→0} sin(1/x): as x → 0, 1/x races to infinity and sin(1/x) oscillates between −1 and 1 forever, never settling.

Reading limits from a graph

[GRAPH: piecewise h(x) on [−1, 7] × [−2, 7] - Smooth curve rising from the left, approaching a HOLE (open circle) at (2, 4) from both sides - A solid dot at (2, 1) marks the actual value h(2) = 1 - The curve continues right and approaches an open circle at (5, 0) from the left - A solid dot at (5, 3) starts the right branch, which rises toward the right edge - Tick marks make heights 0, 1, 3, and 4 readable]

Three readings, three different answers:

When finding a limit, ignore open/closed dots and trace where the curve is heading. Use the dots only when asked for the function's actual value.

Estimating a limit from a table

When substitution fails and you have no graph, squeeze in on c from both sides numerically. The classic (with x in radians):

Both sides march toward 1, so lim_{x→0} (sin x)/x = 1. The dash at 0 is expected — the function is undefined there, but the trend pins down the limit. In Section 2 of the exam, the built-in TI-84's TABLE feature does exactly this job.

Limit laws and direct substitution

If lim_{x→c} f(x) = L and lim_{x→c} g(x) = M both exist:

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     (M ≠ 0)
Power/Root:       lim fⁿ = Lⁿ,  lim ⁿ√f = ⁿ√L  (L ≥ 0 for even n)

The fastest evaluation method is direct substitution: if f is continuous at c (polynomials, rationals with nonzero denominator, roots, exponentials, logs, trig on their domains), then lim_{x→c} f(x) = f(c). Always try substitution first. It fails usefully in one signature way: the indeterminate form 0/0. That is not an answer — it is a signal to do algebra. (By contrast, a form like 5/0 is not indeterminate; it signals unbounded behavior, covered in Lesson 2.)

Resolving 0/0 — three techniques

1. Factor and cancel. When a rational expression gives 0/0, numerator and denominator share a factor of (x − c):

lim_{x→2} (x² + x − 6)/(x − 2) = lim_{x→2} (x + 3)(x − 2)/(x − 2) = lim_{x→2} (x + 3) = 5

Canceling is legal because the limit only cares about x ≠ c.

2. Multiply by a conjugate. When a square root causes the 0/0, use (√A − B)(√A + B) = A − B²:

lim_{x→0} (√(x + 4) − 2)/x = lim_{x→0} x/[x(√(x+4) + 2)] = lim_{x→0} 1/(√(x+4) + 2) = 1/4

Multiply top and bottom by the conjugate, expand only the rooted part, and the offending factor cancels.

3. Combine fractions. For a complex fraction, put the small fractions over a common denominator first:

lim_{x→0} [1/(x + 2) − 1/2]/x = lim_{x→0} (−x)/[2x(x + 2)] = lim_{x→0} −1/[2(x + 2)] = −1/4

The special trig limits

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

To use the first, the argument of sine must match the denominator. lim_{x→0} (sin 5x)/x is not 1 — rewrite it: 5 · (sin 5x)/(5x) → 5 · 1 = 5. Mixed constants work the same way: (sin 7x)/(2x) = (7/2) · (sin 7x)/(7x) → 7/2.

The Squeeze Theorem (recognition level)

If g(x) ≤ f(x) ≤ h(x) near c (except possibly at c) and lim_{x→c} g(x) = lim_{x→c} h(x) = L, then lim_{x→c} f(x) = L.

The CLEP-standard application: a bounded oscillation crushed by a factor going to 0. Since −1 ≤ sin(1/x) ≤ 1,

−x² ≤ x² sin(1/x) ≤ x²   and both outer limits are 0, so  lim_{x→0} x² sin(1/x) = 0.

Contrast this with bare sin(1/x), which has no crushing factor and no limit at 0. The pattern to recognize: (something → 0) × (something bounded) → 0.

Key Takeaways

• A limit describes behavior near a point, never the value at it; limit and function value can disagree.
• A two-sided limit exists only when both one-sided limits exist and agree; otherwise the CLEP answer is nonexistent — caused by a jump, unbounded behavior, or oscillation.
• Try direct substitution first. A result of 0/0 is a signal to factor, rationalize with a conjugate, or combine fractions — then substitute.
• (sin u)/u → 1 only when the same u sits on top and bottom; insert constants to force the match.
• Bounded-times-vanishing → 0 by the Squeeze Theorem: x² sin(1/x) → 0 even though sin(1/x) alone has no limit.
• CLEP items have five choices (A)–(E) with short computational stems; Section 1 is calculator-free, so all of this lesson's algebra must run by hand.

Practice Questions

Questions 1–4 refer to the graph below.

[GRAPH: piecewise h(x) on [−1, 7] × [−2, 7] - Smooth curve approaching a HOLE (open circle) at (2, 4) from both sides - A solid dot at (2, 1) marks h(2) = 1 - The curve continues right and approaches an open circle at (5, 0) from the left - A solid dot at (5, 3) starts the right branch, which rises toward the right edge]

Show answer key & explanations

Answer Key

1. D) 4. Approaching x = 2 from both sides, the curve heads to height 4 (the open circle), so the limit is 4 — the hole is irrelevant to the limit. Distractors: A) 1 is h(2), the function value, not the limit — the classic near-vs-at confusion. B) 2 grabs the x-value instead of a height. C) 0 reads the wrong feature of the graph (the later left-hand limit at x = 5). E) "nonexistent" wrongly assumes a hole kills a limit; both one-sided limits agree at 4, so it exists. Fix: to find a limit from a graph, trace the curve's height as you approach — ignore every dot until a question asks for a function value.

2. A) 1. The solid dot at (2, 1) is the function's actual value: h(2) = 1. Distractors: B) 2 repeats the input x = 2 as if it were the output. C) 4 is the limit at x = 2, not the value — the same trap as Q1 in reverse. D) 5 misreads the jump location x = 5 as a height. E) "undefined" ignores the solid dot; the function is defined there. Fix: solid dot = value, open circle = no value; f(c) questions are dot questions.

3. A) 0. Coming in from the left of x = 5, the curve approaches the open circle at height 0, so the left-hand limit is 0. Distractors: B) 3 is the right-hand limit (the solid dot's branch) — wrong side. C) 5 repeats the x-value. D) 4 reads the height from the wrong location (x = 2). E) "nonexistent" confuses the two-sided limit (which does fail here) with the one-sided limit, which exists on its own. Fix: for x → c⁻, cover the graph to the right of c and read where the remaining curve is heading.

4. E) nonexistent. The left-hand limit is 0 and the right-hand limit is 3; since 0 ≠ 3, the two-sided limit is nonexistent. Distractors: A) 0 uses only the left side. B) 3 uses only the right side (and the solid dot). C) 3/2 illegally averages the two one-sided limits — no averaging rule exists. D) 5 repeats the x-value. Fix: a two-sided limit exists only if left = right; if they differ, the answer is "nonexistent," not either side and not their average.

5. C) 12. From both sides the outputs close in on 12 (11.996 from below, 12.004 from above), so the table pins the limit at 12. Distractors: A) 7 is f(3), the value at the point — the limit ignores it. B) 3 is the input being approached, not an output. D) 11.996 freezes at the last row shown instead of extrapolating the trend. E) "nonexistent" is wrong because both sides agree on the same target. Fix: read a limit table by asking "what single number are both sides sneaking up on?" — never report the point's own value or the last table entry.

6. B) 6. Substitution gives 0/0, so factor: (x² − 9)/(x − 3) = (x − 3)(x + 3)/(x − 3) = x + 3 → 6. Distractors: A) 0 stops at the 0/0 form and calls it 0. C) 3 substitutes into the wrong surviving expression (reports x itself). D) 9 evaluates x² at 3 after mangling the cancellation. E) "nonexistent" treats a removable 0/0 as a failure — the hole fills to a clean limit. Fix: 0/0 in a rational expression means a common factor of (x − c) is hiding: factor, cancel, then substitute.

7. E) 1/8. Multiply by the conjugate: (√(x+16) − 4)/x · (√(x+16) + 4)/(√(x+16) + 4) = x/[x(√(x+16) + 4)] = 1/(√(x+16) + 4) → 1/(4 + 4) = 1/8. Distractors: A) 0 stops at the 0/0 form. B) 1/16 doubles the denominator by squaring instead of adding the two roots. C) 1/4 uses 1/√16 alone, forgetting the + 4 from the conjugate. D) 4 reports √16 itself. Fix: after a conjugate multiply, the finish line is always 1/(√(c + a) + b) — add the root to the constant, then reciprocate.

8. D) −1/25. Combine over a common denominator: 1/(x+5) − 1/5 = (5 − (x+5))/(5(x+5)) = −x/(5(x+5)); dividing by x leaves −1/(5(x+5)) → −1/25. Distractors: A) 1/25 drops the negative from 5 − (x + 5) = −x. B) 0 stops at the 0/0 form. C) −1/5 forgets to keep the factor 5 from the common denominator. E) "nonexistent" misreads a removable 0/0 as failure. Fix: in 1/(x+a) − 1/a problems, the combined numerator is −x; track that minus sign to the final answer.

9. E) 7/2. Force the argument match: (sin 7x)/(2x) = (7/2)·(sin 7x)/(7x) → (7/2)·1 = 7/2. Distractors: A) 0 confuses this with (1 − cos x)/x. B) 1 applies (sin u)/u → 1 without fixing the mismatched constants. C) 2/7 inverts the correction factor. D) 7 corrects the numerator but forgets the 2 downstairs. Fix: (sin ax)/(bx) → a/b — the answer is the sine's coefficient over the denominator's coefficient.

10. C) 0. Since −1 ≤ sin(1/x) ≤ 1, we have −x² ≤ x² sin(1/x) ≤ x², and both bounds → 0, so the Squeeze Theorem forces the limit to 0. Distractors: A) −1 and B) 1 grab the bounds of the sine factor, ignoring the crushing x². D) 1/2 imports an unrelated trig-limit value. E) "nonexistent" applies the fate of bare sin(1/x) — but here the vanishing factor tames the oscillation. Fix: (bounded factor) × (factor → 0) = 0 by the Squeeze Theorem; oscillation alone is fatal only when nothing crushes it.

11. A). sin(1/x) oscillates between −1 and 1 infinitely fast as x → 0 and never settles, so its limit is nonexistent. Distractors: B) x sin(1/x) → 0 — the factor x squeezes it. C) (sin x)/x → 1, the special trig limit. D) (x² − 4)/(x − 2) = x + 2 → 4, a removable 0/0. E) (1 − cos x)/x → 0, the companion trig limit. Each of these exists, so choosing it means misjudging a standard form. Fix: memorize the short list — bare sin(1/x) DNE; squeezed, special-trig, and removable-0/0 forms all exist.

12. B) −4/3. By the limit laws: numerator 2(3) − (−2) = 8, denominator 3·(−2) = −6, so the limit is 8/(−6) = −4/3. Distractors: A) 4/3 loses the negative from the denominator product. C) 3/4 inverts the fraction after computing both parts. D) −2/3 adds g's limit instead of subtracting the negative (numerator 4 instead of 8). E) "nonexistent" wrongly assumes quotients of limits fail — the denominator's limit is −6 ≠ 0, so the quotient law applies. Fix: when both limits exist, plug the limit values straight into the expression; only a denominator limit of 0 blocks the quotient law.