Limits are the foundation of every other idea in calculus. The derivative is a limit. The definite integral is a limit. Continuity, asymptotic behavior, and the major theorems (IVT, MVT, EVT, FTC) all rest on limit reasoning. Roughly is drawn directly from this unit, but the weight is much larger because limits power Units 2โ8 as well.
10โ12 % of the AP Calculus AB exam
real
This page is the unit hub: it gives you the conceptual framing, the AP-style "must-know" skills, and a roadmap of all the granular sub-topics you can study below.
What you'll learn in this unit
What a limit is โ both intuitively (where the function is heading) and formally (the ฮตโฮด definition AP uses informally).
How to compute limits four ways: graphically, numerically (tables), algebraically (factoring, rationalizing, substitution, special trig limits), and via L'Hรดpital's Rule (introduced formally in Unit 4 but previewed here for 0/0 and โ/โ forms).
One-sided limits and what they tell you about jump discontinuities and vertical asymptotes.
Limits at infinity โ end behavior, horizontal asymptotes, and the rational-function rule of thumb.
Infinite limits โ vertical asymptotes from the inside out.
Continuity at a point (three-part definition) and on an interval, and the three flavors of discontinuity (removable, jump, infinite).
Big-picture theorems โ Intermediate Value Theorem, Squeeze Theorem.
The big idea
A limit asks: "As x gets arbitrarily close to a, what value is f(x) getting arbitrarily close to?"
It does not ask what f(a) is. That's the trick: f(a) may be undefined, the wrong value, or anything else, and the limit can still exist. This separation between value at a point and behavior near a point is exactly what lets calculus describe instantaneous rates and exact areas.
Three things a limit can do at x=a
Equal a finite number.limxโaโf(x)=L. The function approaches a single value from both sides.
Equal ยฑโ (an "infinite limit"). The function blows up; this signals a vertical asymptote at x=a.
Fail to exist. Either the left- and right-hand limits disagree (jump), the function oscillates without settling, or ยฑโ disagrees on the two sides.
A limit exists (in the AP sense of "equals a number") only when both one-sided limits agree on a finite value. Any other behavior โ jumps, oscillation, blow-ups โ means the limit DNE.
Computing limits โ the AP playbook
When asked to evaluate limxโaโf(x), follow this order:
Try direct substitution. If f is continuous at a, lim=f(a). Done.
If you get 0/0, look for algebraic simplification. Most common moves:
Factor and cancel (e.g., xโ2x2โ4โ=x+2 for x๎ =2).
Rationalize (multiply by conjugate when there's a square root).
Combine fractions in the numerator.
Use a trig identity (e.g., sin2x+cos2x=1).
Recognize special trig limits: limxโ0โxsinxโ= and .
For limits at infinity of rational functions, compare leading-term degrees:
degree(num) < degree(den): limit = 0
degree(num) = degree(den): limit = ratio of leading coefficients
degree(num) > degree(den): limit is ยฑโ (no horizontal asymptote)
If you still get an indeterminate form (0/0 or โ/โ), use L'Hรดpital's Rule: limgf.
Continuity in one breath
A function f is continuous at x=a iff all three are true:
f(a) is defined.
limxโaโf(x) exists.
limxโaโf(x)=f(a).
If any one fails, f is discontinuous at a. The flavor depends on which one:
Discontinuity
What goes wrong
Fixable by redefining f(a)?
Removable ("hole")
lim exists, but f(a) doesn't equal it (or doesn't exist)
Yes
Jump
Left and right limits exist but disagree
No
Infinite
lim=ยฑโ; vertical asymptote
No
The Intermediate Value Theorem (IVT)
If f is continuous on [a,b] and N is any value between f(a) and f(b), then there is at least one cโ(a,b) with f(c)=N.
In AP problems, the IVT is the go-to justification for "show that f(x)=0 has a solution on [a,b]" or "show that f takes the value 5 somewhere on [a,b]." Always state the continuity hypothesis explicitly when you cite the IVT โ graders will not award the point if you don't.
The Squeeze Theorem
If g(x)โคf(x)โคh(x) near a (except possibly at a) and limxโaโg(x)=limxโaโh(x)=L, then limxโaโf(x)=L.
Most-used template: showing that limxโ0โx2sin(1/x)=0 by sandwiching between โx2 and x2.
How this unit shows up on the AP exam
Multiple choice (no calculator): Algebraic limit evaluations (0/0 form, factoring, rationalizing). Continuity diagnostics from a piecewise definition. Limits-at-infinity / horizontal asymptotes.
Multiple choice (calculator): Estimating limits from a table or graph; verifying a removable discontinuity numerically.
Free response: A continuity argument citing IVT; setting up a piecewise function so it's continuous (solve for a parameter); using one-sided limits to characterize a vertical asymptote.
Common mistakes to avoid
Computing f(a) instead of the limit. They are not the same thing โ the limit ignores the value at a.
Saying "lim=โ" means the limit exists. On the AP exam, an infinite "limit" means the limit does not exist in the formal sense. Use ยฑโ to describe the behavior, not to claim existence.
Skipping the continuity hypothesis when citing IVT. No "continuous on [a,b]" โ no credit.
Plugging โ into rational functions directly. Always compare degrees first.
Forgetting the absolute-value subtlety.limxโ0โxโฃxโฃโ does not exist (left = , right = ).
Quick reference card
Limit exists โ left limit = right limit = same finite number
Indeterminate forms to attack: 0/0, โ/โ (then factor / rationalize / L'Hรดpital)
Special trig: limxโ0โsin(x)/x=1; lim
Rational function at โ: compare leading-term degrees
IVT requires continuity on a closed interval
Sub-topics in this unit
Use the cards below to drill into each granular skill. Start with the conceptual ones (what a limit is, notation, one-sided limits) and move into the algebraic-technique sections (factoring, rationalizing, indeterminate forms) before tackling continuity and limits at infinity. There's also an interactive lesson and entrance quiz at the top of this page that test the whole unit at once.
๐ Practice Problems
1Problem 1easy
โ Question:
Evaluate xโ3limโ(2x2โ5x+1).
๐ก Show Solution
The function is a polynomial โ continuous everywhere โ so the limit equals the value:
limxโ3โ(2x.
2Problem 2easy
โ Question:
Evaluate xโ2limโx.
3Problem 3easy
โ Question:
Given the piecewise function f(x)=, find , , , and . Is continuous at ?
4Problem 4medium
โ Question:
Evaluate xโ0limโx.
5Problem 5medium
โ Question:
Evaluate xโโlimโ and .
6Problem 6medium
โ Question:
Evaluate xโ0limโ3.
7Problem 7medium
โ Question:
Find all values of a that make f(x)= continuous at .
8Problem 8hard
โ Question:
Use the Intermediate Value Theorem to show that f(x)=x3+xโ1 has a root in the interval .
โ ๏ธ Common Mistakes: Limits & Continuity (AP Calculus AB Unit 1)
Avoid these 4 frequent errors
๐ Real-World Applications: Limits & Continuity (AP Calculus AB Unit 1)
See how this math is used in the real world
๐ Worked Example: Related Rates โ Expanding Circle
Problem:
A stone is dropped into a still pond, creating a circular ripple. The radius of the ripple is increasing at a rate of 2 cm/s. How fast is the area of the circle increasing when the radius is 10 cm?
What is Limits & Continuity (AP Calculus AB Unit 1)?โพ
Limit definition, evaluation, one-sided limits, squeeze theorem, and IVT
How can I study Limits & Continuity (AP Calculus AB Unit 1) effectively?โพ
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 10 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Limits & Continuity (AP Calculus AB Unit 1) study guide free?โพ
Yes โ all study notes, flashcards, and practice problems for Limits & Continuity (AP Calculus AB Unit 1) on Study Mondo are free to access. No account is needed.
What course covers Limits & Continuity (AP Calculus AB Unit 1)?โพ
Limits & Continuity (AP Calculus AB Unit 1) is part of the AP Calculus AB course on Study Mondo, specifically in the Limits & Continuity section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Limits & Continuity (AP Calculus AB Unit 1)?โพ
Yes, this page includes 10 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.
1
limxโ0โx1โcosxโ=0
โ
=
limgโฒfโฒโ
โ1
+1
xโ0โ
(
1
โ
cosx)/x=
0
2
โ
5x+
1)=
2(9)โ
5(3)+
1=
18โ
15+
1=
4โ
โ
2
x2โ4
โ
๐ก Show Solution
Direct substitution gives 0/0, an indeterminate form. Factor the numerator:
xโ2x2โ4โ=xโ2(xโ2)(x+2)โ=x+2 for x๎ =2.
So limxโ2โxโ2.
Note that f(2) itself is undefined (the original function has a removable discontinuity at x=2), but the limit exists.
โฉโจโงโ
2x+154โxโx<1x=1x>1โ
limxโ1โโf(x)
limxโ1+โf(x)
limxโ1โf(x)
f(1)
f
x=1
๐ก Show Solution
Left limit: limxโ1โโf(x)=2(1)+1=3.
Right limit: limxโ1+โf(x)=4โ1=.
Since both one-sided limits equal 3, limxโ1โf(x)=3.
f(1)=5 (from the middle piece).
The limit exists and f(1) is defined, but limxโ1โf(x)๎ =. Continuity fails the third condition โ . The discontinuity is (redefining would fix it).
x+9โโ3
โ
๐ก Show Solution
Direct substitution gives 0/0. Multiply numerator and denominator by the conjugate x+9โ+3:
x.
Now plug in x=0: 9.
2x2+xโ43x2โ5x+7
โ
xโโlimโx2โ24x+1โ
๐ก Show Solution
Rule of thumb (rational function at โ): compare leading-term degrees.
(a) Top and bottom both have degree 2. Limit = ratio of leading coefficients = 3/2โ.
(b) Top has degree 1, bottom has degree 2. Bottom grows faster, so the ratio โ0โ. (The line y=0 is a horizontal asymptote.)
x
sin(5x)
โ
๐ก Show Solution
Use the special trig limit limuโ0โusinuโ=1.
Rewrite: 3xsin(5x)โ=5x.
As xโ0, 5xโ0, so 5x.
lim=1โ 35โ=.
{ax+3x2โ1โxโค2x>2โ
x=2
๐ก Show Solution
Continuity at x=2 requires the two pieces to agree there:
Left value: f(2)=2a+3.
Right limit: limxโ2+โf(x)=22โ1=3.
Set equal: 2a+3=3โ2a=0โa=0.
(0,1)
๐ก Show Solution
Step 1.f(x)=x3+xโ1 is a polynomial โ continuous on [0,1] (and everywhere). The IVT continuity hypothesis is satisfied.
Step 2. Compute the endpoints:
f(0)=03+0โ1=โ1
Step 3.f(0)=โ1<0<1=f(1). So 0 lies between and .
Step 4. By the IVT, there exists cโ(0,1) with f(c)=0. โ
AP grading note: explicitly stating "continuous on [0,1]" is required for full credit.
x2+4x
โ
โ
x
)
๐ก Show Solution
Direct substitution gives โโโ, indeterminate. Multiply by the conjugate:
x2+4xโโx=.
Divide top and bottom by x (with x>0 so x2):
x2+4x.
As xโโ, 4/xโ0:
lim=1โ+1.
2
sin
(x1โ)
๐ก Show Solution
Step 1 โ Bound the sine. For all x๎ =0, โ1โคsin(1/x)โค1.
Step 2 โ Multiply by x2 (positive for x๎ =0):.
Step 3 โ Take limits of the outer functions:limxโ0โ(โx2)=0 and .
Step 4 โ Apply the Squeeze Theorem:0โคlimxโ0โx2sin(1/x)โค, so the limit equals .
Note that sin(1/x) alone has no limit at 0 (it oscillates wildly), but multiplying by x2 damps the oscillation.