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Quizzes > Mathematics

Limits and Continuity Quiz: Check Your Calculus Foundations

Quick, free calculus continuity test. Instant results.

Editorial: Review CompletedCreated By: Lee CallaghanUpdated Aug 25, 2025
Difficulty: Moderate
2-5mins
Learning OutcomesCheat Sheet
Cut paper art math symbols and function graphs for calculus continuity quiz on golden yellow background

Use this limits and continuity quiz to check one-sided limits, continuity at a point, and typical discontinuities. Get instant results to spot weak areas before a test. Keep practicing with a calculus limits quiz, review rules in a derivative rules quiz, or build skills with an integration quiz.

Which of the following best describes continuity of a function f(x) at x = a?
The limit from the left equals the limit from the right, and both equal f(a).
The function f(x) is differentiable at x = a.
The limit of f(x) as x approaches a exists.
The value f(a) is defined.
A function is continuous at x = a if the left-hand limit, right-hand limit, and the value of the function at that point all coincide. This ensures there is no jump or hole at x = a. Differentiability is a stronger condition than continuity, and existence of the limit or function value alone is not sufficient. Learn more about continuity definitions .
Which statement is true regarding polynomial functions?
They are continuous for all real numbers.
They are only continuous on closed intervals.
They have vertical asymptotes where the degree is odd.
They have discontinuities at their zeros.
Polynomial functions are sums and products of continuous power functions, so they are continuous everywhere on the real line. They have no breaks, jumps, or vertical asymptotes. Zeros of a polynomial do not imply discontinuities. More on polynomial continuity .
What is the value of lim??5 (3x + 2)?
17
16
15
13
A polynomial limit can be evaluated by direct substitution: 3(5) + 2 = 15 + 2 = 17. There is no discontinuity or indeterminate form in a linear function. This is a straightforward application of limit properties. See more .
Given f(x) = {(x² - 4)/(x - 2) if x ? 2; 5 if x = 2}, is f continuous at x = 2?
Yes, because the simplified function equals 4.
Yes, because f(2) = 4.
No, because the function is not defined at x = 2.
No, because the limit from both sides is 4, which does not equal f(2) = 5.
The expression simplifies to x + 2 for x ? 2, so the limit as x approaches 2 is 4. Since f(2) is defined as 5, the limit does not equal the function value, creating a removable discontinuity. A removable discontinuity occurs when a hole can be filled but isn't. More details .
What type of discontinuity does f(x) = (x² - 4)/(x - 2) have at x = 2?
Infinite discontinuity
Removable discontinuity
Oscillatory discontinuity
Jump discontinuity
Since (x² - 4)/(x - 2) simplifies to x + 2 except at x = 2, the function has a hole at x = 2 that can be removed by redefining the value. This is the hallmark of a removable discontinuity. There are no jumps or vertical asymptotes here. See an explanation .
Is the function f(x) = { x if x < 0; -x if x ? 0 } continuous at x = 0?
No, it has a cusp at 0.
No, there is a hole at x = 0.
Yes, because both one-sided limits and f(0) are 0.
No, it has a jump discontinuity at 0.
For x < 0, f(x) = x approaches 0 from the left; for x ? 0, f(x) = -x approaches 0 from the right, and f(0) = 0. Since both one-sided limits equal the function value, f is continuous at 0. This is typical for the absolute value function. More on piecewise continuity .
Which condition is necessary for the Intermediate Value Theorem to apply to f on [a, b]?
f is differentiable on (a, b).
f is monotonic on [a, b].
f is continuous on [a, b].
f is bounded on [a, b].
The Intermediate Value Theorem states that if f is continuous on a closed interval [a, b], then it takes every value between f(a) and f(b). Differentiability, boundedness, or monotonicity are not required for the IVT. Continuity on the entire interval is the key hypothesis. Read more .
On which interval is g(x) = ?x continuous?
(-?, 0)
(-?, ?)
[0, ?)
(0, ?)
The square root function is defined and continuous for all x ? 0. It is not defined for negative x in the real number system. On its domain [0, ?), the function has no breaks or jumps. More details .
The function f(x) = sin(1/x) for x ? 0 and f(0) = 0 is:
Discontinuous at x = 0
Continuous but not differentiable at x = 0
Has a removable discontinuity at x = 0
Continuous for all real x
As x approaches 0, sin(1/x) oscillates between -1 and 1 and does not settle to a single value, so the limit does not exist. Since the limit is not equal to f(0), the function is discontinuous at 0. This is an example of an essential discontinuity. See more .
What type of discontinuity does f(x) = {1 if x < 0; 2 if x ? 0} have at x = 0?
Oscillatory discontinuity
Removable discontinuity
Infinite discontinuity
Jump discontinuity
The left-hand limit as x approaches 0 is 1, while the right-hand limit is 2, so the function value jumps from 1 to 2 at x = 0. This is the definition of a jump discontinuity. There is no hole or vertical asymptote here. More details .
Which type of discontinuity occurs at x = 1 for h(x) = (x² + 1)/(x² - 1)?
Jump discontinuity
Removable discontinuity
Oscillatory discontinuity
Infinite discontinuity
At x = 1 the denominator is zero while the numerator is nonzero, creating a vertical asymptote. This is characteristic of an infinite discontinuity. There is no hole to remove because the limit diverges. Learn more .
The greatest integer function f(x) = ?x? is discontinuous at which points?
All integers
All rational numbers
All irrational numbers
All non-integers
The floor function jumps down by 1 at each integer value from the right, so there is a jump discontinuity at every integer. Between integers, the function is constant and continuous. See more .
For f(x) = 2x, to prove continuity at x = 3 using the ?-? definition, one must choose ? as:
?/3
?
?/2
2?
We require |2x - 6| < ?, which simplifies to 2|x - 3| < ?, so |x - 3| < ?/2. Thus choosing ? = ?/2 guarantees the definition holds. This is the standard ?-? proof method. See an example .
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Study Outcomes

  1. Understand Continuity Criteria -

    Explain the formal definition of continuity and identify the three conditions required for a function to be continuous at a given point.

  2. Apply Limit Tests -

    Use various limit evaluation techniques to verify continuity or detect discontinuities in algebraic and piecewise functions.

  3. Analyze Types of Discontinuities -

    Differentiate among removable, jump, and infinite discontinuities and classify them in the context of real and rational functions.

  4. Evaluate Continuity in Context -

    Interpret continuity in real-world scenarios by assessing function behavior near critical points and practical applications.

  5. Diagnose Weak Spots -

    Pinpoint common misconceptions and challenging areas in continuity to focus your review and strengthen problem-solving skills.

  6. Master Fundamental Continuity Rules -

    Recall and apply key theorems such as the Intermediate Value Theorem to solve continuity questions in calculus confidently.

Cheat Sheet

  1. Formal Definition of Continuity at a Point -

    Continuity at x=a requires lim x→a f(x)=f(a), which implies both one-sided limits agree with the function value. This three-part criterion, highlighted in Stewart's Calculus (8th ed.), is crucial for tackling any test for continuity calculus.

  2. Classifying Discontinuities -

    Identify removable (holes in the graph), jump (sudden value shifts), and infinite discontinuities (vertical asymptotes) by inspecting limits from each side. For example, f(x)=(x²−1)/(x−1) has a removable discontinuity at x=1, while f(x)=1/(x−2) has an infinite discontinuity at x=2.

  3. Intermediate Value Theorem -

    The IVT states that if f is continuous on [a,b] and k is between f(a) and f(b), there exists c in (a,b) such that f(c)=k, a fact often tested in limits and continuity quizzes. Remember the mnemonic "no breaks, no gaps" to recall that continuous functions hit every intermediate value.

  4. Extreme Value Theorem -

    On a closed interval [a,b], a continuous function achieves both a global maximum and minimum, a principle you'll likely see on a calculus continuity practice test. This guarantees that optimization problems on [a,b] have attainable extrema.

  5. One-Sided Continuity at Endpoints -

    At the endpoints of a domain, check only the relevant one-sided limit: lim x→a❺ f(x)=f(a) or lim x→b❻ f(x)=f(b), vital for continuity questions in calculus on closed intervals. For instance, f(x)=√x is continuous at x=0 because lim x→0❺ √x=0.

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