Calculus I · Limits and Continuity · practice
Continuity and IVT Practice Problems
Visual study stop
Read the picture before the symbols
Pause at each graph long enough to say what the input is doing, what the output is doing, and which feature supports the next mathematical decision.
Read this graph as text
Four types of discontinuity. A two-by-two gallery compares behavior at x = 0. Removable shows y = x + 1 with an open point at (0, 1). Jump shows y = 0 to the left and y = 2 to the right with open and filled markers. Infinite shows y = 1/x on split domains with a vertical asymptote. Oscillatory shows y = sin(1/x) on split domains that never settle near zero.
Panel titles name each type. Open and filled markers distinguish inclusion, the jump branches use different line styles, and the infinite panel has a dashed asymptote.
Why it matters: Compare removable, jump, infinite, and oscillatory discontinuities in a consistent four-panel layout.
A removable hole, jump, vertical blow-up, and oscillation fail continuity for different reasons. Only a removable mismatch can be repaired by changing one function value.
Read this graph as text
A root guaranteed by the Intermediate Value Theorem. The continuous curve f(x) = x cubed + x - 1 is shown on the closed interval from 0 to 1. A filled circle at (0, -1) lies below the x-axis and a filled square at (1, 1) lies above it. The curve crosses the axis at a filled diamond c approximately 0.6823, illustrating a root whose existence the Intermediate Value Theorem guarantees.
The negative endpoint is a filled circle, the positive endpoint is a filled square, and the root is a filled diamond, all with text labels.
Why it matters: Show how continuity and opposite endpoint signs guarantee at least one zero between the endpoints.
Opposite endpoint signs and continuity guarantee a root in the interval. They do not locate it exactly or prove that it is the only root.
Section 5 Exercises
A. Three-part continuity test
Is continuous at ? Verify all three conditions.
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Yes; value and limit both equal .
Answer 1 from the source-traced unit appendix.Is continuous at ? Identify the first failed condition.
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No; the function value is undefined and behavior is unbounded.
Answer 2 from the source-traced unit appendix.A function has and . Is it continuous at ?
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Yes.
Answer 3 from the source-traced unit appendix.A function has and . Which condition fails?
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The equality condition fails.
Answer 4 from the source-traced unit appendix.A function is undefined at , but its limit there is . Which conditions fail?
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The function-value condition fails, and therefore the equality condition cannot hold.
Answer 5 from the source-traced unit appendix.A function has unequal one-sided limits at . Can it be continuous there?
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No.
Answer 6 from the source-traced unit appendix.Explain why existence of alone says nothing about continuity.
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A dot may be isolated from the nearby graph.
Answer 7 from the source-traced unit appendix.Explain why existence of a finite limit alone does not guarantee continuity.
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The value may be missing or different from the limit.
Answer 8 from the source-traced unit appendix.B. Intervals of continuity
Find intervals of continuity of .
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Answer 9 from the source-traced unit appendix.Find intervals of continuity of .
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Answer 10 from the source-traced unit appendix.Find intervals of continuity of .
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Answer 11 from the source-traced unit appendix.Find intervals of continuity of .
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Answer 12 from the source-traced unit appendix.Find intervals of continuity of on .
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Break at ; continuous on the resulting subintervals.
Answer 13 from the source-traced unit appendix.Find intervals of continuity of .
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Answer 14 from the source-traced unit appendix.Find the domain and intervals of continuity of .
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Answer 15 from the source-traced unit appendix.Explain endpoint continuity for on .
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Continuous on , using right continuity at and left continuity at .
Answer 16 from the source-traced unit appendix.C. Classifying discontinuities
Classify the discontinuity of at .
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Removable.
Answer 17 from the source-traced unit appendix.Classify the discontinuity of at .
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Jump.
Answer 18 from the source-traced unit appendix.Classify the discontinuity of at .
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Infinite.
Answer 19 from the source-traced unit appendix.Classify the discontinuity of at .
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Oscillatory.
Answer 20 from the source-traced unit appendix.Find and classify every discontinuity of .
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Hole at ; vertical asymptote at .
Answer 21 from the source-traced unit appendix.Find and classify every discontinuity of .
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Infinite discontinuity at after one factor cancels.
Answer 22 from the source-traced unit appendix.Can changing one function value repair a jump? Explain.
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No; nearby one-sided behavior remains unequal.
Answer 23 from the source-traced unit appendix.Can changing one function value repair a vertical asymptote? Explain.
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No; nearby values remain unbounded.
Answer 24 from the source-traced unit appendix.D. Repairing holes
Find so makes continuous.
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Answer 25 from the source-traced unit appendix.Find so makes continuous.
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Answer 26 from the source-traced unit appendix.Find so makes continuous.
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Answer 27 from the source-traced unit appendix.Find so makes continuous.
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Answer 28 from the source-traced unit appendix.Explain why no value at makes continuous there.
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No value works because the one-sided limits are and .
Answer 29 from the source-traced unit appendix.Explain why no value at makes continuous there.
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No value works because the function is unbounded.
Answer 30 from the source-traced unit appendix.E. Piecewise parameters
Find so is continuous at .
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Answer 31 from the source-traced unit appendix.Find so is continuous at .
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Answer 32 from the source-traced unit appendix.Find so is continuous at .
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Answer 33 from the source-traced unit appendix.Find so is continuous at .
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Answer 34 from the source-traced unit appendix.Determine whether any makes continuous at .
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No value works.
Answer 35 from the source-traced unit appendix.Find so is continuous at both joins.
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Answer 36 from the source-traced unit appendix.Design a piecewise function with one parameter that becomes continuous when the parameter is .
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Answers vary.
Answer 37 from the source-traced unit appendix.Explain why setting left and right formulas equal is necessary but not always sufficient when the actual function value is defined by a third rule.
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The third rule must assign the common limiting value at the join.
Answer 38 from the source-traced unit appendix.F. Intermediate Value Theorem and bisection
Show that has a root in .
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Polynomial continuity and signs at .
Answer 39 from the source-traced unit appendix.Show that has a root in .
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Polynomial continuity and signs at .
Answer 40 from the source-traced unit appendix.Show that has a solution in .
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is continuous; , .
Answer 41 from the source-traced unit appendix.Explain why a sign change is sufficient but not necessary for a root to exist.
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A function can touch zero and return without an endpoint sign change.
Answer 42 from the source-traced unit appendix.Give a continuous function with a root in whose endpoint values have the same sign.
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Example: on .
Answer 43 from the source-traced unit appendix.Explain why IVT cannot be applied to on , despite opposite endpoint signs.
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is not continuous on the interval because it is undefined at zero.
Answer 44 from the source-traced unit appendix.Use two bisection steps to narrow a root of from .
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After two steps, the root lies in .
Answer 45 from the source-traced unit appendix.Use three bisection steps to narrow a root of from .
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After three steps, the root lies in .
Answer 46 from the source-traced unit appendix.Does IVT prove uniqueness? Give an example supporting your answer.
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No. Example: has two roots on .
Answer 47 from the source-traced unit appendix.State every hypothesis and conclusion in a complete IVT solution.
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State continuity on , an intermediate target between endpoint outputs, and the existence of with .
Answer 48 from the source-traced unit appendix.Answers begin in the referenced section.
After the explanation
Use the section idea
Do the limit, the function value, and the surrounding domain fit together at the point or across the interval?
Continuity is a three-part agreement: the value exists, the two-sided limit exists, and those two quantities are equal.
At a point, test the three conditions in order; on an interval, check the domain and endpoints before invoking any continuity theorem.
A sign change supports the Intermediate Value Theorem only when continuity holds on the entire closed interval, and it does not prove uniqueness.
You are ready to continue when you can classify a break, decide whether one value can repair it, and state every IVT hypothesis aloud.
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