Study guide: Introductory Calculus: Limits and Continuity
A limit describes nearby approach behavior rather than necessarily a function’s value at the point. One-sided agreement, removable holes and continuity connect algebraic simplification with precise statements about a function’s local behavior.
Ideas to explain in your own words
- A rational expression may simplify for every nearby input except its missing point. The simpler rule then describes approach behavior, allowing a limit even when the original function is undefined there. Cancellation is valid only away from the zero denominator.
- A two-sided limit exists only when both one-sided limits exist and agree. A function’s separately assigned value at the point cannot reconcile different left and right approach values. Do not average the two one-sided limits or substitute the isolated point’s value.
- Continuity at a point requires the function value to be defined, the limit to exist, and that value to equal the limit. A removable hole can be filled with its approach value to meet these conditions. Changing a point’s value can fix a removable hole but cannot fix unequal one-sided limits.
- Dividing numerator and denominator by x reveals reciprocal terms that approach zero as x grows without bound. With a nonzero denominator leading coefficient, the remaining coefficient ratio determines this limit. Infinity is an approach direction rather than a number to substitute.
Useful vocabulary
- limit
- A value approached by function outputs as inputs approach a location.
- one-sided limit
- Approach behavior from only the left or only the right.
- continuity
- A defined function value matching the existing limit at a point.
Watch your reasoning
- Cancellation is valid only away from the zero denominator. A computed limit does not fill the original domain hole by itself; write the distinction between nearby values and the value at the point.
- Do not average the two one-sided limits or substitute the isolated point’s value. Compare the approach values directly; different values imply that the two-sided limit does not exist.
- Changing a point’s value can fix a removable hole but cannot fix unequal one-sided limits. State all three continuity conditions and retain the distinction between the original rule and the newly defined extension.
A worked review example
Complete supplied guide examples: Find an Approach Value across a Removable Hole — Find lim as x→2 of (x²−4)/(x−2). Candidate answers: (A) 3 (B) 4 (C) 5 (D) 8. For x≠2, factor and cancel to get x+2. As x approaches 2, this approaches 4. The original function is still undefined exactly at the excluded point. Require Left and Right Approaches to Agree — Near x=0, f(x)=2 for x<0 and f(x)=3 for x>0; f(0)=12. Find the two-sided limit at 0. Candidate answers: (A) The limit is 12 (B) The limit is 2 (C) The limit is 3 (D) The two-sided limit does not exist. The left-hand limit is 2; the right-hand limit is 3. They differ, so no two-sided limit exists. Changing f(0) would not reconcile those approach values. Choose a Value that Matches the Limit — For x≠2, f(x)=(x²−4)/(x−2); define f(2)=k. What k makes f continuous at 2? Candidate answers: (A) 3 (B) 4 (C) 5 (D) 8. The nearby simplified rule is x+2, whose limit at 2 is 4. Continuity requires the defined function value to equal that limit, so k=4. Compare Leading Linear Terms at Large Inputs — Find lim as x→∞ of (2x+1)/(3x−2). Candidate answers: (A) -1 (B) 0 (C) 2/3 (D) 3/2. Divide numerator and denominator by x: (2+1/x)/(3−2/x). The reciprocal terms approach zero, leaving 2/3. The denominator approaches nonzero 3.
Make a practice plan
- Study the four methods in the listed order; attempt each independent check before opening its explanation.
- Complete each lesson’s three changed-input questions and record the operation or reasoning step behind any correction.
- Take the four-question unit review, then solve the transfer task with written justifications.
- After correcting errors, attempt the separate unit exam and revisit the linked lesson for any unresolved method.