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📝 Continuity definition in calculus (18 MCQs)

📖 From Calculus • 2. Limits and Continuity an Introduction • 18 questions available

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What is Continuity definition in calculus?

Definition:
A function $f$ is continuous at a point $a$ if three conditions hold: $f(a)$ is defined, $\lim_{x \to a} f(x)$ exists, and $\lim_{x \to a} f(x) = f(a)$ . This means there is no break, jump, or hole at $a$ , and the function's value matches its limiting behavior, ensuring that small changes in input produce small changes in output, which is the essence of a smooth, connected graph.

Example:
Check continuity of $f(x) = x^2$ at $x=2$ .
Solution: $f(2)=4$ , $\lim_{x \to 2} x^2 = 4$ , so $\lim = f(2)$ ; hence continuous.

Reason:
Continuity is fundamental because it ensures functions behave predictably, allowing for the application of important theorems like the Intermediate Value Theorem, and is a prerequisite for differentiability and integrability in calculus.

Easy

Medium

Hard

📝 All Continuity definition in calculus MCQs

Q1. Which of the following correctly states the definition of continuity of a function f at a point c?

A.f(c) is defined, the limit as x→c exists, and limₓ→c f(x)=f(c). ✅

B.f(c) is defined and differentiable at c.

C.Only the limit as x→c exists.

D.Only f(c) is defined.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The definition requires three facts: the function value at c must exist, the two‑sided limit as x approaches c must exist, and the limit must equal the function value. All three together guarantee no break, hole, or jump at the point, which is precisely the formal notion of continuity.

Q2. If limₓ→c f(x) exists and equals L, but f(c)≠L, the discontinuity at c is called:

A.Jump discontinuity

B.Infinite discontinuity

C.Removable discontinuity ✅

D.Essential discontinuity

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: When the limit exists and is finite but differs from the actual function value, the interruption can be “removed” by redefining the function at that point. This situation is known as a removable discontinuity because the graph has a hole that can be filled without altering the surrounding behavior.

Q3. Consider f(x)=⎧ x² for x<1; 2 for x=1; x+2 for x>1⎫. Which condition of continuity fails at c=1?

A.f(1) is undefined

B.The limit as x→1 does not exist

C.The limit exists but ≠f(1) ✅

D.All conditions are satisfied

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: At x=1 the left‑hand limit is 1²=1, the right‑hand limit is 1+2=3, so the two‑sided limit does not exist. Because the limit fails to exist, the third condition (limit equals function value) cannot hold, indicating a discontinuity.

Q4. If a function has a jump discontinuity at c, which statement must be false?

A.The left‑hand limit equals the right‑hand limit ✅

B.Both one‑sided limits exist

C.The overall limit as x→c does not exist

D.The function value at c may be defined or not

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A jump discontinuity means the left‑hand and right‑hand limits approach different numbers; therefore they cannot be equal. The other statements can be true: each one‑sided limit can exist, the two‑sided limit fails to exist, and the function’s value at c is irrelevant to the jump.

Q5. Suppose limₓ→c f(x)=5 while f(c) is undefined. What can be concluded about continuity at c?

A.f is continuous at c

B.f has a removable discontinuity at c

C.f has an infinite discontinuity at c ✅

D.No discontinuity exists at c

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: When the limit exists but the function value is missing, the first condition of continuity fails. Since the limit is finite, the discontinuity is removable; redefining f(c) to equal the limit would restore continuity. Hence the situation represents a removable discontinuity.

Q6. For a function with an infinite discontinuity at c, which description fits its one‑sided limits?

A.Both approach finite but unequal numbers

B.Both diverge to +∞ or −∞

C.At least one does not exist as a real number ✅

D.Both equal the function value at c

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: An infinite discontinuity occurs when at least one one‑sided limit grows without bound (to +∞ or −∞) or fails to approach any real number. Consequently, the limit as x→c does not exist in the real sense, distinguishing it from jump or removable cases.

Q7. If limₓ→c f(x)=f(c)=5 and f is continuous everywhere else, what does this imply about the graph near c?

A.There is a hole at c

B.The graph passes smoothly through (c,5) without breaks ✅

C.There is a vertical asymptote at c

D.The function is undefined at c

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: When the limit equals the function value and all three continuity conditions hold, the curve has no interruption at that point. The graph therefore continues uninterrupted through the coordinate (c,5), showing a smooth, unbroken passage that reflects true continuity.

Q8. Consider f(x)=\frac{x^{2}-4}{x-2} and g(x)=x+2. Which statement about their continuity at x=2 is true?

A.Both are continuous at 2

B.f is discontinuous, g is continuous

C.f becomes continuous after simplification, g is discontinuous ✅

D.Both are discontinuous at 2

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The expression f(x) simplifies to x+2 for x≠2, so the limit as x→2 exists and equals 4. However, f is undefined at 2, causing a removable discontinuity. The function g(x)=x+2 is a polynomial, hence continuous everywhere, including 2.

Q9. Which of the following functions is continuous at x=0?

A.h(x)=\sin x ✅

B.k(x)=\frac{x}{|x|}

C.m(x)=\frac{x^{2}}{x}

D.n(x)=\sqrt{x}

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The sine function is defined for all real numbers and its limit as x→0 equals 0, matching its value at 0, so it is continuous. The other options either have a domain restriction (√x requires x≥0) or a jump at 0 (x/|x|) or are undefined at 0 (x²/x).

Q10. Which statement correctly distinguishes a removable discontinuity from a jump discontinuity?

A.Removable has unequal one‑sided limits, jump has equal limits

B.Removable limit exists and is finite; jump limit does not exist ✅

C.Both have infinite limits

D.Neither can be fixed by redefining f(c)

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A removable discontinuity occurs when both one‑sided limits exist and equal the same finite number, yet the function value differs. A jump discontinuity features two finite one‑sided limits that are unequal, so the overall limit fails to exist. This contrast captures the essential difference.

Q11. Between a function with a removable discontinuity and one with a jump discontinuity at the same point, which typically exhibits a larger visual gap in its graph?

A.Removable discontinuity

B.Jump discontinuity ✅

C.Both have equal gaps

D.Neither has a gap

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: A jump discontinuity creates a sudden break where the left‑hand and right‑hand limits differ, producing a noticeable vertical gap. A removable discontinuity, by contrast, is merely a single missing point (a hole) that can be patched by redefining the function value, so the visual interruption is much smaller.

Q12. To make a function f continuous at c where it currently has a removable discontinuity, which condition must the new definition f(c)=? satisfy?

A.Equal the left‑hand limit

B.Equal the right‑hand limit

C.Equal the common limit of both sides ✅

D.Any finite number

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For a removable discontinuity the two one‑sided limits already agree; the limit exists as a single number L. Restoring continuity requires assigning f(c) exactly that common limit L, ensuring that the third continuity condition (limit equals function value) holds.

Q13. Why is every polynomial p(x) continuous at any real number c?

A.Polynomials are differentiable

B.Polynomials are built from continuous operations

C.Polynomials have no limits

D.Polynomials are bounded ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: Polynomials are formed by adding, subtracting, and multiplying constant and power functions, each of which is continuous everywhere. The composition of continuous functions remains continuous, so any polynomial inherits continuity at every real point without exception.

Q14. If a function f is continuous on [a,b] and f(a)=f(b), which theorem guarantees a point c∈(a,b) where f'(c)=0?

A.Intermediate Value Theorem

B.Mean Value Theorem

C.Rolle's Theorem ✅

D.Extreme Value Theorem

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Rolle's Theorem is a special case of the Mean Value Theorem that applies when the function values at the interval endpoints are equal. It asserts the existence of at least one interior point where the derivative vanishes, reflecting the horizontal tangent required by continuity and differentiability.

Q15. How does continuity of a function on [a,b] ensure the Intermediate Value Theorem holds?

A.It forces the function to be monotonic

B.It prevents jumps, so every value between f(a) and f(b) is attained ✅

C.It guarantees differentiability

D.It creates a vertical asymptote at each endpoint

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Continuity eliminates any breaks or holes in the graph, meaning the function cannot skip over any intermediate values. Consequently, for any number between the endpoint values, the graph must cross that height somewhere in the interval, which is precisely the guarantee provided by the Intermediate Value Theorem.

Q16. A function f has a removable discontinuity at c because limₓ→c f(x)=L but f(c)=M≠L. Which action will restore continuity?

A.Redefine f(c)=L

B.Remove the point c from the domain

C.Change the limit to M ✅

D.Add a vertical asymptote at c

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Since the limit exists and is finite, the only obstacle to continuity is the mismatched function value. By redefining f(c) to equal the limit L, the third condition of continuity is satisfied, eliminating the hole and making the function continuous at c.

Q17. If limₓ→c f(x)=4 exists, yet f(c)=7, what type of discontinuity is present and how can it be corrected?

A.Jump; redefine to 4

B.Removable; redefine to 4

C.Infinite; remove point c ✅

D.Essential; cannot fix

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The situation matches a removable discontinuity: the limit exists (4) but the function value differs (7). Correcting it simply requires redefining f(c) to equal the limit 4, thereby satisfying all three continuity conditions and eliminating the discontinuity.

Q18. Let f(x)=⎧ \frac{x^{2}-1}{x-1} for x<0; 2 for x=0; \sqrt{x} for x>0⎫. Is f continuous at x=0?

A.Yes, because both one‑sided limits equal 2

B.No, because the left‑hand limit is 1 while the right‑hand limit is 0, so the overall limit does not exist ✅

C.No, because f(0) is undefined

D.No, because there is a vertical asymptote at 0

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: For x<0, the expression simplifies to x+1, giving a left‑hand limit of 1 as x→0⁻. The right‑hand limit from √x approaches 0 as x→0⁺. Since the one‑sided limits differ, the two‑sided limit does not exist, violating the second continuity condition; thus f is discontinuous at 0.

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