k that makes a piecewise function continuous — a standard CLEP numeric-entry item.A function f is continuous at x = c exactly when all three of these hold:
(1) f(c) is defined
(2) lim_{x→c} f(x) exists (left-hand limit = right-hand limit, both finite)
(3) lim_{x→c} f(x) = f(c) (the limit equals the value)
If any one condition fails, f is discontinuous at c. The order matters: (2) needs the one-sided limits to agree; (3) then needs that common value to match the height of the point.
The friendly function families — polynomials, rational functions (where the denominator is nonzero), root, exponential, logarithmic, and trigonometric functions — are continuous at every point of their domains. Sums, products, quotients (nonzero denominator), and compositions of continuous functions stay continuous. So for these you evaluate a limit simply by substituting.
lim_{x→c} f(x) exists as a finite number, but f(c) is undefined or unequal to it. You could "patch" the hole by redefining f(c). These come from a factor that cancels: (x²−9)/(x−3) = x+3 for x ≠ 3, hole at (3, 6).+∞ or −∞, typically at a vertical asymptote such as 1/(x−c).[GRAPH: three panels on [-1, 5] × [-4, 4]
- REMOVABLE: line y = x+3 with an open circle (hole) at (3, 6) — "limit exists, value missing"
- JUMP: segment at height 1 ending in an open circle at (2,1); segment at height 3 starting with a filled dot at (2,3) — "left limit 1 ≠ right limit 3"
- INFINITE: y = 1/(x-2) with a dashed vertical asymptote at x = 2, branches diving to -∞ and rising to +∞]
A classic CLEP item gives a piecewise function with an unknown constant and asks you to make it continuous. Each piece is already continuous on its own, so the only place to check is the boundary. Force the two pieces to agree there.
Example. For
f(x) = x² + kwhenx ≤ 2andf(x) = 3xwhenx > 2, findksofis continuous atx = 2. - Left piece at 2:(2)² + k = 4 + k- Right limit asx → 2⁺:3(2) = 6- Set equal:4 + k = 6, sok = 2.
When f(x) grows without bound near c, the limit is +∞ or −∞, and the line x = c is a vertical asymptote. Watch the sign on each side:
lim_{x→2} 1/(x−2)² = +∞ (denominator is positive on BOTH sides, → 0)
lim_{x→0⁻} 1/x = −∞ (just left of 0, x is a small negative number)
lim_{x→0⁺} 1/x = +∞ (just right of 0, x is a small positive number)
±∞ describes how a limit fails; it is not a finite value.
Ask what happens as x → +∞ or x → −∞. If f(x) → L (finite), then y = L is a horizontal asymptote. For a rational function, compare the degrees of numerator and denominator:
| Degree comparison | lim_{x→±∞} |
Horizontal asymptote |
|---|---|---|
| numerator degree < denominator degree | 0 |
y = 0 |
| numerator degree = denominator degree | ratio of leading coefficients | y = a/b |
| numerator degree > denominator degree | ±∞ (no finite limit) |
none |
lim_{x→∞} (3x² − 5x)/(2x² + 1) = 3/2 (equal degree, 3/2 = ratio of leading coefficients)
lim_{x→∞} (5x + 1)/(x² + 4) = 0 (bottom wins)
lim_{x→∞} (2x³ + 1)/(x² − 5) = +∞ (top wins)
IVT. If
fis continuous on the closed interval[a, b]andNis any number strictly betweenf(a)andf(b), then there is at least onecin the open interval(a, b)withf(c) = N.
The IVT is an existence theorem: it guarantees some c, without telling you where it is or how many there are. Both hypotheses are load-bearing — continuity on a closed interval, and N genuinely between the endpoint values. To show a root exists, take N = 0 and confirm a sign change: for f(x) = x³ + x − 1, f(0) = −1 < 0 < 1 = f(1), so a root lies in (0, 1).
c requires all three: f(c) defined, the limit exists, and they are equal.±∞ one-sided limit.0; equal degree → leading-coefficient ratio; top-heavier → ±∞.N strictly between the endpoint values; it only guarantees existence.f(x) = (x² − 9)/(x − 3) has what type of discontinuity at x = 3?1. E. Correct: x² − 9 = (x − 3)(x + 3) cancels the (x − 3), so the limit is 6 but f(3) is undefined — a removable hole. A) jump needs unequal finite one-sided limits. B) infinite needs a factor that does not cancel. C) nothing oscillates. D) f(3) is undefined, so not continuous. Fix rule: if the zero-denominator factor cancels, the discontinuity is removable.
k is f(x) = x² + k for x ≤ 2 and f(x) = 3x for x > 2 continuous at x = 2?2. C. Correct: at the boundary 4 + k = 6, so k = 2. A) 0 ignores the boundary equation. B) 1 mis-solves. D) 4 uses only the left constant term. E) 6 copies the right-side value without subtracting 4. Fix rule: set left piece = right piece at the boundary and solve.
lim_{x→∞} (3x² − 5x)/(2x² + 1) is3. C. Correct: equal degrees, so the limit is the ratio of leading coefficients 3/2. A) 0 would need a bottom-heavier fraction. B) 2/3 inverts the ratio. D) ∞ would need a top-heavier fraction. E) the limit exists. Fix rule: equal degree → leading-coefficient ratio.
lim_{x→2} 1/(x − 2)² is4. E. Correct: (x − 2)² is positive on both sides and shrinks to 0, so 1/(x − 2)² grows to +∞. A) 0 misreads the growth. B) −∞ wrong sign — the square is never negative. C) 1 ignores the blow-up. D) both sides agree at +∞, so this is the described behavior. Fix rule: even-power denominators give the same sign from both sides.
f is continuous on [1, 5] with f(1) = −3 and f(5) = 7. For which value N does the IVT guarantee some c in (1, 5) with f(c) = N?5. D. Correct: the IVT delivers every value strictly between f(1) = −3 and f(5) = 7; only 2 lies in (−3, 7). A) −5 and E) −4 are below −3. B) 9 and C) 8 exceed 7. Fix rule: N must be strictly between the two endpoint outputs.
g satisfies lim_{x→3⁻} g(x) = 5 and lim_{x→3⁺} g(x) = 1. The discontinuity at x = 3 is6. B. Correct: both one-sided limits are finite but unequal (5 and 1) — a jump. A) removable needs a two-sided limit to exist. C) infinite needs an unbounded side. D) unequal limits mean not continuous. E) nothing oscillates. Fix rule: finite but unequal one-sided limits = jump.
f(x) = (x + 2)/(x² − 4) has an infinite discontinuity at7. A. Correct: (x + 2)/[(x − 2)(x + 2)] = 1/(x − 2); the (x + 2) cancels (removable at −2) but (x − 2) does not, so x = 2 is infinite. B) x = −2 is removable, not infinite. C) x = 0 is in the domain. D) only x = 2 is infinite. E) an infinite discontinuity does exist. Fix rule: the non-cancelling zero-denominator factor gives the infinite discontinuity.
f(x) = (4x³ − x)/(2x³ + 7) is8. B. Correct: equal degree 3, ratio of leading coefficients 4/2 = 2, so y = 2. A) y = 0 needs a bottom-heavier fraction. C) 1/2 inverts the ratio. D) an equal-degree rational function does have a horizontal asymptote. E) 4/7 misuses the constant terms. Fix rule: horizontal asymptote of equal-degree rational = leading-coefficient ratio.
f is continuous at x = c. Which of the following CANNOT be true?9. A. Correct: continuity at c forces lim_{x→c} f(x) = f(c), so A directly contradicts continuity and cannot happen. B) a continuous function may equal 0. C) continuity does not require differentiability (e.g., |x|). D) a horizontal tangent is compatible with continuity. E) this is exactly what continuity asserts. Fix rule: continuous ⇒ limit equals value.
f(x) = (x² − x − 6)/(x − 3) for x ≠ 3 and f(3) = k is continuous at x = 3 when k =10. D. Correct: (x² − x − 6)/(x − 3) = x + 2 for x ≠ 3, so the limit is 5; setting k = 5 matches it. A) 0 ignores the limit. B) 3 uses the x-value. C) 6 mis-evaluates x + 2 at x = 3 as if x + 3. E) the limit exists, so a value works. Fix rule: the patch value equals the (cancelled) limit.
lim_{x→∞} (2x³ + 1)/(x² − 5) is11. D. Correct: numerator degree 3 exceeds denominator degree 2, so the ratio grows without bound: +∞. A) 0 would need bottom-heavier. B) 2 treats the degrees as equal. C) 1/2 inverts. E) the behavior is well described as ∞. Fix rule: top-heavier rational → ±∞.
f(x) = x³ + x − 1, note f(0) = −1 and f(1) = 1. The Intermediate Value Theorem guarantees12. E. Correct: f is a polynomial (continuous), and f(0) = −1 < 0 < 1 = f(1), so by the IVT there is at least one c in the open interval (0, 1) with f(c) = 0. A) the IVT never counts roots. B) the sign change forces a root. C) f(0) = −1 ≠ 0. D) the IVT gives no upper count. Fix rule: a sign change plus continuity guarantees at least one root in the open interval.
1. E. Correct: x² − 9 = (x − 3)(x + 3) cancels the (x − 3), so the limit is 6 but f(3) is undefined — a removable hole. A) jump needs unequal finite one-sided limits. B) infinite needs a factor that does not cancel. C) nothing oscillates. D) f(3) is undefined, so not continuous. Fix rule: if the zero-denominator factor cancels, the discontinuity is removable.
2. C. Correct: at the boundary 4 + k = 6, so k = 2. A) 0 ignores the boundary equation. B) 1 mis-solves. D) 4 uses only the left constant term. E) 6 copies the right-side value without subtracting 4. Fix rule: set left piece = right piece at the boundary and solve.
3. C. Correct: equal degrees, so the limit is the ratio of leading coefficients 3/2. A) 0 would need a bottom-heavier fraction. B) 2/3 inverts the ratio. D) ∞ would need a top-heavier fraction. E) the limit exists. Fix rule: equal degree → leading-coefficient ratio.
4. E. Correct: (x − 2)² is positive on both sides and shrinks to 0, so 1/(x − 2)² grows to +∞. A) 0 misreads the growth. B) −∞ wrong sign — the square is never negative. C) 1 ignores the blow-up. D) both sides agree at +∞, so this is the described behavior. Fix rule: even-power denominators give the same sign from both sides.
5. D. Correct: the IVT delivers every value strictly between f(1) = −3 and f(5) = 7; only 2 lies in (−3, 7). A) −5 and E) −4 are below −3. B) 9 and C) 8 exceed 7. Fix rule: N must be strictly between the two endpoint outputs.
6. B. Correct: both one-sided limits are finite but unequal (5 and 1) — a jump. A) removable needs a two-sided limit to exist. C) infinite needs an unbounded side. D) unequal limits mean not continuous. E) nothing oscillates. Fix rule: finite but unequal one-sided limits = jump.
7. A. Correct: (x + 2)/[(x − 2)(x + 2)] = 1/(x − 2); the (x + 2) cancels (removable at −2) but (x − 2) does not, so x = 2 is infinite. B) x = −2 is removable, not infinite. C) x = 0 is in the domain. D) only x = 2 is infinite. E) an infinite discontinuity does exist. Fix rule: the non-cancelling zero-denominator factor gives the infinite discontinuity.
8. B. Correct: equal degree 3, ratio of leading coefficients 4/2 = 2, so y = 2. A) y = 0 needs a bottom-heavier fraction. C) 1/2 inverts the ratio. D) an equal-degree rational function does have a horizontal asymptote. E) 4/7 misuses the constant terms. Fix rule: horizontal asymptote of equal-degree rational = leading-coefficient ratio.
9. A. Correct: continuity at c forces lim_{x→c} f(x) = f(c), so A directly contradicts continuity and cannot happen. B) a continuous function may equal 0. C) continuity does not require differentiability (e.g., |x|). D) a horizontal tangent is compatible with continuity. E) this is exactly what continuity asserts. Fix rule: continuous ⇒ limit equals value.
10. D. Correct: (x² − x − 6)/(x − 3) = x + 2 for x ≠ 3, so the limit is 5; setting k = 5 matches it. A) 0 ignores the limit. B) 3 uses the x-value. C) 6 mis-evaluates x + 2 at x = 3 as if x + 3. E) the limit exists, so a value works. Fix rule: the patch value equals the (cancelled) limit.
11. D. Correct: numerator degree 3 exceeds denominator degree 2, so the ratio grows without bound: +∞. A) 0 would need bottom-heavier. B) 2 treats the degrees as equal. C) 1/2 inverts. E) the behavior is well described as ∞. Fix rule: top-heavier rational → ±∞.
12. E. Correct: f is a polynomial (continuous), and f(0) = −1 < 0 < 1 = f(1), so by the IVT there is at least one c in the open interval (0, 1) with f(c) = 0. A) the IVT never counts roots. B) the sign change forces a root. C) f(0) = −1 ≠ 0. D) the IVT gives no upper count. Fix rule: a sign change plus continuity guarantees at least one root in the open interval.