Calculus I · Limits and Continuity · exam
Limits and Continuity Practice Exam B
Answer key published
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The complete key is online, numbered to match this exam, and linked to the verified source appendix.
Practice Examination B
This exam is deliberately less patterned. It is meant to test whether you can identify methods rather than imitate section headings.
Suggested time: 90 minutes.
Let
Find both one-sided limits, the two-sided limit, , and determine continuity at .
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Left , right , two-sided limit , function value ; not continuous, removable discontinuity.
Answer 1 from the source-traced unit appendix.Show answer
Cancel : .
Answer 2 from the source-traced unit appendix.Show answer
Rationalization gives .
Answer 3 from the source-traced unit appendix.Show answer
.
Answer 4 from the source-traced unit appendix.Show answer
.
Answer 5 from the source-traced unit appendix.Show answer
Squeeze gives .
Answer 6 from the source-traced unit appendix.Find every discontinuity of , classify it, and state all one-sided infinite limits.
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Hole at ; vertical asymptote . Both one-sided limits at are .
Answer 7 from the source-traced unit appendix.Show answer
.
Answer 8 from the source-traced unit appendix.Find the polynomial asymptote of the function in the previous problem.
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Division gives polynomial asymptote .
Answer 9 from the source-traced unit appendix.Show answer
.
Answer 10 from the source-traced unit appendix.Find so
is continuous at and .
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Continuity at gives , so ; continuity at then gives , so .
Answer 11 from the source-traced unit appendix.Show that has a root in , then perform two bisection steps.
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The polynomial is continuous, , . First midpoint is negative, so interval . Second midpoint is positive, so interval .
Answer 12 from the source-traced unit appendix.Prove using the formal definition.
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Choose ; then .
Answer 13 from the source-traced unit appendix.A student claims that because , every function has a root in . Give a counterexample and state the missing hypothesis.
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Counterexample: on . Endpoint signs differ, but the function is not continuous on the interval. The missing hypothesis is continuity.
Answer 14 from the source-traced unit appendix.Full answers and selected worked solutions are in the appendices. Do not read them until you have completed an honest attempt. The universe already contains enough answer keys masquerading as education.
After the explanation
Use the section idea
Can you diagnose the limit type and justify a method before beginning the algebra?
A mixed problem is a classification task before it is a calculation: direction, substitution result, structure, and required conclusion determine the route.
Name the limit type and first legal move in a margin note, then solve and check whether the conclusion matches the graph or sign behavior.
Pattern matching without diagnosis makes similar-looking problems blur together and hides whether the error was conceptual, algebraic, or strategic.
You are exam-ready when you can choose a method without a section label, explain the choice, and correct a miss by naming its exact cause.
Source & rights
Original instruction with traceable references.
The exposition is original. No Active Calculus exercise is reproduced verbatim. Public-domain examples were modernized and recomposed when used as inspiration.
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